Mathematical embarrassments are problems that should be solved already

Terry Tao is one of the greatest mathematicians of our time. Tao has already solved many long-standing open problems, which earned him a Fields Medal. What I cannot understand about him is how he can be so productive. He is the author of one of the best blogs in the world, he writes original articles that solve hard open problems, he writes great survey articles on a wide range of topics, and he is the author of terrific books on topics from additive combinatorics to partial differential equations. He is impressive.

Today I want to talk about a topic that he has discussed: mathematical embarrassments—the name is mine, but I hope he will agree with spirit of this term.

There is an old story about Godfrey Hardy and John Littlewood, who together, of course, published many great papers. Steven Krantz tells a version of the story in his fun book, Mathematical apocrypha:

It is said that Landau thought “Littlewood” was a pseudonym for Hardy so that it would not seem like he wrote all the papers.

I think Tao might consider using this trick and invent a co-author or two. It would help make the rest of us feel better.

Let’s now turn to mathematical embarrassments.

Mathematical Embarrassments

A mathematical embarrassment (ME) is a problem that should have been solved by now. An ME usually is easy to state, seems approachable, and yet resists all attempts at an attack. There may be many reasons that they are yet to be solved, but they “feel” like they should have been solved already.

Some Examples

${\bullet }$ The ${\pi}$ and ${e}$ problem. The problem is to prove that ${\pi+e}$ and ${\pi-e}$ are both transcendental numbers. We know that one of these must be transcendental. For if both were algebraic, then so would,

$\displaystyle \frac{1}{2}(\pi + e + \pi -e) = \pi,$

which contradicts the known fact that ${\pi}$ is transcendental.

Similarly, at least one of ${\pi + e}$ and ${\pi e}$ must be transcendental, for otherwise

$\displaystyle x^2 - (\pi + e) x + \pi e$

would be an algebraic polynomial with transcendental roots. It seems ridiculous that we are able to show the transcendence of ${\pi}$ and ${e}$, but not these numbers.

${\bullet }$ The linear recursion zero problem. An ME due to Tao is a problem about linear recurrences. Given a linear recurrence over the integers,

$\displaystyle a_{n} = c_{1}a_{n-1} + \dots + c_{d}a_{n-d}$

with initial conditions, does there exist an ${k}$ so that ${a_{k} = 0}$? He only asks for a decision procedure: he says that it is ${\dots}$ faintly outrageous that this problem is still open; it is saying that we do not know how to decide the halting problem even for “linear” automata!

${\bullet }$ Mersenne composites. The problem is to prove that for an infinite number of primes ${p}$,

$\displaystyle 2^{p} - 1$

is composite. It is an open question whether there are infinite number of Mersenne primes. Currently, only 47 Mersenne primes are known. It seems more likely that there are an infinite number of Mersenne composites. Leonhard Euler has shown that

Theorem: If ${k>1}$ and ${p=4k+3}$ is prime, then ${2p+1}$ is prime if and only if ${2^p = 1 \bmod{2p+1}}$.

Thus, if ${p = 4k + 3}$ and ${2p+1}$ are both primes, then ${2^p - 1}$ is a composite number. So an approach to the problem is to show there are infinitely many ${p}$‘s of the above form.

${\bullet }$ The Jacobian conjecture. This is one of my favorite problems outside theory proper. Consider two polynomials ${f(x,y)}$ and ${g(x,y)}$. The Jacobian Conjecture asks when is the mapping:

$\displaystyle (x,y) \rightarrow (f(x,y),g(x,y))$

one to one? The values of ${x,y}$ are allowed to be any complex numbers. Clearly, a simple necessary condition is that the mapping be locally one to one: this requires that the determinant of the Jacobian of the mapping to be everywhere non-zero. That is that for all ${x}$ and ${y}$,

$\displaystyle \det \begin{pmatrix} f_{x}(x,y) & g_{x}(x,y) \\ f_{y}(x,y) & g_{y}(x,y) \end{pmatrix} = c$

for some non-zero constant ${c}$. If the value of ${c}$ is not a constant, it is a non-trival polynomial, which must take the value zero for some values ${x',y'}$. So for those values, the mapping will not even be one to one locally.

Note, ${f_{x}(x,y)}$ is the value of the partial derivative of ${f}$ with respect to ${x}$, and the same for ${y}$. The Jacobian Conjecture is that this necessary condition is sufficient.

This is the conjecture for two dimensions; there is a similar one for higher dimensions, but this case is already open. Note, the maps can be messy; for example, the mapping

$\displaystyle (x,y) \rightarrow (x+y^{7}, y)$

is one to one. Then, compose this with some linear invertible transformation, and the mapping can start to look very strange.

The Jacobian Conjecture is related to another famous theorem called the Ax—Grothendieck theorem. In fact, the Jacobian conjecture implies the Ax—Grothendieck theorem.

Theorem: Suppose that ${f:\mathbb{C}^{n} \rightarrow \mathbb{C}^{n}}$ is a polynomial function. Then, if ${f}$ is one-to-one, then ${f}$ is onto.

The proof is based on the following trivial observation: if a map from a finite set to itself is one-to-one, then it is onto. Amazing. The full proof is based on a simple compactness argument and this simple observation. See this for a nice exposition by Michael O’Connor.

${\bullet }$ The existence of explicit non-linear lower bounds. One of the great ME for theory is that we have no circuit lower bounds that are non-linear on any explicit problem. The natural proof and other barriers do not seem to be relevant here. I do not believe that they stop a lower bound of the form:

$\displaystyle n\log^{*} n.$

${\bullet }$ The BPP hierarchy problem. The problem is to show that ${\mathsf{BPTIME}(n^{100})}$ is more powerful than ${\mathsf{BPTIME}(n)}$. Even with respect to oracles, we do not know if this is true. It is hard to believe that such a basic question is still open. There is a partial result due to Jin-Yi Cai, Ajay Nerurkar, and D. Sivakumar.

${\bullet }$ The number of planar graphs. How many labeled planar graphs are there, with a given number of vertices ${n}$? If ${P(n)}$ is the number of such graphs, then the best known bounds are of the form:

$\displaystyle n! c_{1}^{n} \le P(n) \le n! c_{2}^{n}$

where ${c_{1} < c_{2}}$. How can we not know the number of planar graphs?

Well it turns out that I am embarrassed—Omer Giménez and Marc Noy know. They have proved—in 2008—an asymptotic estimate for the number of labeled planar graphs:

$\displaystyle g \cdot n^{-7/2}\gamma^{n}n!$

where ${g}$ and ${\gamma}$ are explicit computable constants. This a beautiful result, which solves a long standing ME.

There more known about 2-connected planar graphs—see the paper by Edward Bender, Zhicheng Gao, and Nicholas Wormald for some pretty results. One that I like very much is that they can show that a random 2-connected planar graph is highly likely to contain many copies of any fixed planar graph:

Theorem: For any fixed planar graph ${H}$, there exist positive constants ${c}$ and ${\delta}$ such that the probability that a random labeled 2-connected planar graph G with ${n}$ vertices has less than ${cn}$ vertex disjoint copies of ${H}$ is ${O({e}^{-\delta n})}$.

${\bullet }$ The Barrington problem. David Barrington’s famous theorem shows that polynomial size bounded width computations can compute all of ${\mathsf{NC}^{1}}$. He does this by using computations over simple groups. An open question is can simple groups be replaced by solvable groups? I think the conventional wisdom is that this must be impossible. But there seems to be no progress on this problem. The reason it is interesting is that the construction of Barrington fails easily for solvable groups, but that is not a proof.

See this set of lecture notes and the blog post by Lance Fortnow for more discussion on Barrington’s theorem.

${\bullet }$ The ${n^{d/2}}$ problem. This problem is given a series of integers

$\displaystyle x_{1}, \dots, x_{n}$

is there a subset of size ${d}$ that sums to ${0}$? How can order ${n^{d/2}}$ be the best possible running time?

This algorithm is essentially the same as the exponential algorithm for the knapsack problem that I talked about before. Recall, that the knapsack problem is to find a 0-1 vector ${x}$ so that

$\displaystyle a_{1}x_{1} + \cdots + a_{n}x_{n} = b.$

The method re-writes the problem as

$\displaystyle a_{1}x_{1} + \cdots + a_{m}x_{m} = b - (a_{m+1}x_{m+1} + \cdots + a_{n}x_{n})$

where ${m=n/2}$—we can assume that ${n}$ is even. Then, all the left hand sums are computed, and all the right hand sums are computed too. There is a solution to the original problem if and only if these sets have a value in common. Since there are ${2^{n/2}}$ of them, this method takes time ${2^{n/2}}$ times a polynomial in ${n}$.

${\bullet }$ The Hilbert subspace problem. Given a Hilbert space and a linear operator ${A}$ is there a subspace ${S}$ so that it is non-trivial, invariant under ${A}$, and closed? Invariant under ${A}$ means that for each ${x}$ in ${S}$, ${Ax}$ is also in ${S}$. This is a classic result for finite Hilbert spaces, but is has long been open in general. Several special cases of this problem have been resolved, but not the general case.

Open Problems

What are your favorite ME? Let me know, please.

106 Comments leave one →
1. Jeremy H permalink
December 26, 2009 10:11 am

An unconditional hierarchy has also been proven for BPP/1. See http://portal.acm.org/citation.cfm?id=1272505

2. A mathematical embarrassment permalink
December 26, 2009 11:16 am

It might not be the best idea in the world to title a blog post “Mathematical Embarrassments” and then put a prominent picture of Terry Tao. Usually when there’s a title X and then a picture Y it is inferred that X refers to Y.

December 26, 2009 12:52 pm

I mean no such thing. Sorry for that.

December 26, 2009 11:26 am

The Aanderaa–Karp–Rosenberg conjecture for randomised decision trees perhaps? Since the deterministic Omega(n^2) bound was proven so quickly, but the randomised bound has been open for 30+ years.

December 26, 2009 12:53 pm

Yes. One of the coolest questions. Definitely should have been solved by now.

4. December 26, 2009 12:16 pm

Frankl’s union-closed families conjecture seems like it should be the simplest thing in the world.

5. December 26, 2009 1:24 pm

The d-SUM problem (your “n^{d/2} problem”) actually has a running time of n^ceiling(d/2). A lg^2 n factor improvement is known, but not more. Many people in geometry believe n^ceiling(d/2) / lg^f(d) n is optimal, at least for any constant d. (But we do not have a lg^f(d) upper bound yet, just lg^2…)

Some weak evidence for this is in a recent paper of mine with Ryan Williams: a solution in time n^o(d) would violate the Exponential Time Hypothesis. I do believe that a solution in less than n^ceiling(d/2) time would also imply CNF-SAT can be solved better than 2^n, but proving it is seems beyond current techniques (especially wrt sparsification of SAT formulas).

December 26, 2009 1:28 pm

Yes, I should have pointed to your paper. I plan on discussing it in the future—I enjoyed your paper very much.

• December 26, 2009 5:48 pm

Glad to hear that Do you mean the upper or the lower-bound paper?

December 26, 2009 5:30 pm

We think P = BPP, but we don’t know if BPP is a strict subset of NEXP.

That’s pretty embarassing.

(Potentially, the polynomial hierarchy collapses to BPP = PSPACE. Then we can also have PSPACE = EXP = NEXP.)

7. December 26, 2009 6:14 pm

Why is the Barrington problem interesting? What consequences would there be if you could prove Barrington’s theorem using a different group?

• August 24, 2010 2:56 pm

A somewhat belated reply, I’m afraid: If you could compute all of NC^1 with poly-length programs over a solvable group, then there would be poly-size, constant-depth, unbounded-fanin circuits with AND, OR, and MOD-m gates for any problem in NC^1. (Here m would be an integer depending on the solvable group you used.) So far we haven’t been able to imagine how a constant number of AND, OR, and modular counting operations could solve a problem like MAJORITY. If it could, we would have an example of a new way to compute with these operations.

I think of the problem as how to show that you _can’t_ do all of NC^1 with problems over a solvable group. This would contribute to our understanding because we would then know something about everything you could do with the circuits above (the class “ACC” or “ACC^0″). This has been the “obvious next step” in circuit complexity lower bounds now for over twenty years, which gives the problem further cachet just from its antiquity.

December 26, 2009 6:37 pm

Aren’t mathematical embarrasments pretty much the same things as mathematical diseases?
Both are easy to state and are probably underestimated in their level of difficulty.

9. December 26, 2009 6:55 pm

It is of course very flattering to be mentioned here, but I should point out that the linear recursion problem is not due to myself, and is usually attributed to Skolem, see

http://tucs.fi/publications/insight.php?id=tHaHaHiKa05a

I myself learned of it from Kousha Etassami.

10. December 26, 2009 7:08 pm

The Jacobian conjecture and the invariant subspace problem are also well-known mathematical diseases, though their incidence rate is lower than that for, say, the twin prime conjecture, due to the somewhat more advanced nature of mathematics required even to understand the problem, let alone to believe that one can solve it.

At the risk of propagating these diseases, though, here is a sub-problem of the invariant subspace problem which I find interesting: is the invariant subspace problem finitisable, in the sense that there is a statement about transformations on finite-dimensional Hilbert spaces which is equivalent to the ISP? (I am vague here as to what kinds of statements qualify, but the type of thing one is looking for is that a finite-dimensional transformation should contain a “constructible” “approximate subspace” which is “approximately invariant”, though I do not know how to quantify these concepts precisely.) I looked at this issue some time ago and it appears that there is not enough compactness to make a finitary equivalent form of the ISP, but it could be that a clever rephrasing of the ISP problem could resolve this issue.

11. Michael Mitzenmacher permalink
December 26, 2009 9:39 pm

My personal favorite for the last several years — what is the capacity of the binary deletion channel?

December 26, 2009 11:04 pm

The gap between the upper bounds and lower bounds for the parameters of the triangle removal lemma.

13. December 27, 2009 3:05 am

A true mathematical embarrassement in convex polytope theory is the following question of Imre Barany

Is it true that for every convex polytope P the number of k dimensional faces is at least the minimum of the number of vertices and the number of facets. (Facets are faces of top dimension.)

14. December 27, 2009 12:27 pm

I am at a loss. I am not suggesting any ME; I just had a question.

When does a problem become a mathematical embarrassment from a mathematical disease?

I understand there was an earlier post on MD’s and now ME’s. Are they not the same?

December 27, 2009 3:11 pm

Nope. MDs can be problems that look hard, and feel like they’re going to be really tough to solve. MEs on the other hand sound like problems that should have been solved a long time ago. Perhaps the first time you hear about the ME, you might even be surprised that it hasn’t been solved. There are problems that are in both categories, of course. Perhaps if too many years pass, and very many mathematicians work hard on a MD, and it still remains open, it might become a ME.

15. December 27, 2009 12:54 pm

I would classify the Riemann Hypothesis, for instance, as a disease but not an embarrassment; certainly it is famous and has caused many mathematicians at varying levels of ability to become obsessed with it, but it does not look like a problem which is embarrassingly easy to solve.

Conversely, the Skolem problem about linear recurrences is an embarrassment but not a disease; I would understand if we did not understand the behaviour of iterates a nonlinear system (such as the 3x+1 system), but linear systems, really, should be completely understood by now. On the other hand the problem is somewhat obscure and does not have a reputation for wasting the time of many mathematicians.

16. December 27, 2009 2:51 pm

The $\zeta(5)$ problem. The problem is to prove that $\zeta(5)$ is irrational. In spite of the relative high number of attempts (*) nothing is known about the arithmetic nature of it (or in general of $\zeta(2n+1)$, for n>1). Maybe it is better defined as a mathematical disease.
(*) See e. g. this list compiled by Wadim Zudilin

17. Anindya De permalink
December 27, 2009 4:17 pm

Failure to prove any strong direct product theorems for poly sized circuits or similar strong computational models may count as one.

18. Wei Yu permalink
December 28, 2009 2:32 am

The 3n+1 problem:
http://en.wikipedia.org/wiki/Collatz_conjecture

19. December 28, 2009 3:31 am

Nearest neighbour search on n points in d dimensions still seems to be stuck at $\Theta (2^d \log n)$ running time since one of the very first papers on it by Friedman, Bentley, and Finkel back in the 70′s. There are a ton of approximations around, but nothing exact seems to have been found that does any better, and I don’t think there’s been any proof that $\Theta (2^d \log n)$ is optimal.

There are a ton of applications involving high-dimensional nearest-neighbour search and related problems, and it’s such a simple problem, but getting past the “curse of dimensionality” is always a pain it seems.

20. Matt E permalink
December 28, 2009 1:25 pm

I second Wei Yu–the Collatz conjecture. Maddening!

21. December 28, 2009 4:53 pm

All these are excellent problems.

But yet another, perhaps, “most embarrassed” problem is this.
Its statement requires almost no mathematics.

For a (0,1) matrix A, let t(A) be the smallest number t such that
one can write A as a sum (over the reals) of t 0-1 matrices A_1,…,A_t
such that the 0-entries in each of the A_i can be covered by at most t
all-0 submatrices of A_i.

Easy counting shows that nXn matrices A with t(A)>n^{1/2} exist.

Problem: Exhibit an EXPLICIT matrix A with t(A)>n^c for an arbitrary small
constant c>0.

Be it so “simply,” this is a more than 30 years old problem in computational
complexity! Such a matrix would give us the first super-linear lower bound
on the size of log-depth circuits. The first real progress in this field.

Is this problem indeed harder than classical ones, mentioned above? If so, then why? Just because of “explicit”? Can mathematics *exhibit* interesting objects, not just prove their existence?

December 28, 2009 6:49 pm

Great example.

22. December 28, 2009 5:22 pm

Although the problems in these posts are wonderful, I will submit that none of them are truly embarrassing. The reason is simple: they all are in the category “Problems that look easy, but actually are hard.”

Isn’t it true, that much more embarrassing are mathematical problems that look extremely hard, but actually are easy?

Applied mathematics provides several examples: public key cryptography, fast Fourier transforms, and even (recently) sparse reconstruction algorithms.

In applied physics too, magnetic resonance imaging could have been discovered decades earlier.

What other math/physics problems, that today are generally perceived to be hard (mumpsimus), will someday be perceived as embarrassingly easy (sumpsimus)?

• December 28, 2009 6:44 pm

I agree with you. Disease-like problem should be “look easy, but actually very hard” and embarrassing problem should be “look extremely hard, bu actually very easy”. When a mathematical problem, say a conjecture, known to be either disease-like or embarrassing problem one may attempt to transform it into a “look easy and solve easy” type problem. My two unsuccessful attempts on a disease/embarrassing problem is the graceful tree conjecture (GTC) (see On Mathematical Diseases and More on Mathematical Diseases).
My first attempt was in 1980 when I have transformed the GTC into the (stronger) ordered graceful tree conjecture (OGTC), I thought either an counterexample would be easily found or a breakthrough for the proof of the both conjectures. As the recent works have been shown these two conjectures remain a disease [1],[2],[3].
The second attempt (1986) is to transform the GTC into the simplest form i.e, cordial or two-equitable labeling of trees and try to settle it step-by-step [4],[5]. Unfortunately k-equitability of trees remain a difficult conjecture for all k greater than 3.

[1] I. Cahit,”Another formulation of the graceful tree conjecture”, Research Report CORR 80-47, Dept C&O, University of Waterloo, Ontario, Canada, 1980.
[2] I. Cahit,”On graceful trees”, Bull. Inst. Combin. Appl., 12 (1994) 15-18.
[3] P. Hajnal and G. Nagy, Simply sequentially additive labelings of 2-regular graphs”,
Discrete Mathematics ,310 (4), February 2010, pp. 922-928.
[4] I. Cahit, Cordial graphs : a weaker version of graceful and harmonious graphs, Ars Combin., 23 (1987) 201-207.
[5] I. Cahit, On cordial and 3-equitable labelings of graphs, Utilitas Math.,37 (1990) 189-198.

23. December 28, 2009 7:56 pm

I think that the following question, in the spirit of Fagin’s theorem, is an embarrassment in that it has been open since the birth of descriptive complexity theory: is there an NP or co-NP graph property that is not expressible as an existential second-order formula with a single binary existentially quantified predicate?

I.e., is there an NP or co-NP graph property P not expressible in the following way?

G = (V, E) has property P if and only if
(Exists Q) Phi(Q, V, E),
where Q is a binary relation on the set V of vertices in G and Phi is a first-order formula.

December 30, 2009 1:44 pm

It is proven that almost all real numbers have the property that the geometric mean of the coefficients of their continued fraction expansion is equal to one number, known as Khinchin’s constant. However, we have yet to find a specific example of a number that has the Khinchin property.

Also, I still think of Goldbach’s conjecture as a ME.

• December 31, 2009 9:09 am

Take a look at this paper, where this problem is solved.

Title: A Khinchin Sequence
Author(s): Thomas Wieting
Journal: Proc. Amer. Math. Soc. 136 (2008), 815-824.

• January 20, 2010 1:38 pm

As far as I can tell, Wieting’s paper shows that a particular sequence C is a Khinchin sequence, which is significant progress. However, the associated Khinchin number is only implicitly defined, as the number that has C as its continued fraction representation. The paper notes that no more directly specified numbers are known to be Khinchin numbers, even though almost all irrationals in the interval (0,1) are Khinchin numbers. So I think this remains an interesting “easy-looking, but hard” problem (though personally I’m not sure it qualifies as a ME, with either definition).

January 2, 2010 3:58 pm

Another one: we don’t know if Euler–Mascheroni constant is irrational, not to mention transcendential.

January 8, 2010 3:29 am

What a great post! Later this year I will give a talk about mathematical embarrassments for math teachers. I might add the following two problems:

1. Are the digits of $$\pi$$ normal?
2. Zaremba’s conjecture: for every integer $$q$$ there exists an integer $$p$$ (relatively prime with q) such that the continued fraction $$[a_1,a_2,\dots,a_n]$$ of $$\frac{p}{q}$$ has $$a_i< 5$$ for $$i=1,\dots,n$$.

27. John C permalink
January 14, 2010 10:15 am

How about an elegant readable proof of the four colour theorem?

Made all the more embarrassing because the theorem has already been proved by other means.

28. February 24, 2010 5:50 am

Why prove anything? Why prove the Mersenne primes or Fermats Last theorem? Why prove the fundamental theorem of integral calculus? Why work on the roof when you havent finished building the foundation? The fact is, we have never proven basic arithmetic in full… we have that whole incompleteness bit. What of logic? Modus ponens. How do we prove the foundations of logic and reason without the fallacy of circular reasoning, of assuming the consequent, of begging the question? How do we prove logic without using logic? And all the rest of mathematics that sit atop? We take so much on shear FAITH. We have faith that one apple and one apple makes two apples. We have faith in the notion of non-contradiction. Axioms and definitions are not proven… and calling it an axiom is all you need to do to alleviate the obligation to prove. Just look at the number pi. Its defined as the ratio between circumference and diameter – in euclidean space it is approximately 3.14159, and it is a constant; but what of hyperspace? I heard once that pi is 2 in spherical space. The 3.14159 value for pi is the foundations of trigonometry and calculus and enters into every facet of higher math… and yet its value is entirely based on the notion of Euclidean space – an assumption clearly proven wrong by Einstein. Sure, you can call Euclidean space an axiom requiring no proof if you wanted to… but shouldnt that assumption be demonstrably true, self-consistent and non-contradictory? Our system of math from trig on up is obviously based on a false assumption from trivial geometry. And we turn a blind eye to this inconsistency. Is it any wonder we cant wrap our minds around these difficult unsolved problems?

• February 10, 2012 12:13 am

Right but one has only to know-and not to believe in- the epistemology of existence of anything,here: math, then, the meaning of the term “one” as assigned a sign “1″, the sign” +” (and “-”,….)- read a little of Aristotle@St.Thomas Aquinas@modern mathematicians in this trend, then paradoxes arise from forgetting about this “origin” of math; “being”-”one”-”truth” convertutur (Lat),I.e., are the same in meaning but in different aspects.
Of course, any induction is based on…..believe (or faith) that what happened in n-cases will happen in n+1 cases;ex., the fact that today is A.D.2012 is not a prove that it will continue in 2013, 2014,…. or that the Sun rose up until today for millions of years it will not give us a knowledge it will do tomorrow- ex.G.Boole, St.James Letter in Bible. ONe must decide which word to use for such cases: a predicate ” to know” or “to believe” and consequently to follow in use after that!
Paradoxes arise just from playing on …symbols ( a syntax) and forgetting about their meaning (semantics)-see: ex.A.Tarski, R.Carnap
We know (absolutely) about it and not only believe it – in meta-language level

29. September 4, 2010 2:27 am

There is one question who embarass at least myself, and surprise the few searcher I asked it:

Let \Sigma_j EXP be the class of language decidable in time 2^{n^{O(1)}} begining in existantial state and alternating between existential and universal states at most j-1 times.

If \Sigma_j EXP=\Sigma_{j+1} EXP then can we prove that \Sigma_j EXP=\Sigma_{j+2} EXP ? Like for the polynomial hierarchie, if two level are equals, does the exponential hiearchie collapses ? It seems it is clear than it is true; that they can not be two levels equals, and a third one different… But it seems no one know how to prove it.

In fact the question seems to be open for any time hierarchie bigger than the polynomials.

• September 4, 2010 10:02 am

Lane Hemaspaandra (then surnamed Hemachandra) showed that a related exponential time hierarchy actually does collapse to its second level. His paper “The Strong Exponential Time Hierarchy Collapses” also has general technical machinery relevant to your query—here’s the original Cornell TR version.

• September 5, 2010 3:14 am

Thank you for the link, I knew the result about the strong exponential hierarchy but did not know this paper. It is indeed interesting.
But as he explains in his “open problem” section, this hierarchy does not seems to be equivalent to any Alternating Turing Machine definition.

He also stated that for some classes, if there is a difference, then certains condition must be satisfied. But I do not see in this paper where he proves that it is indeed enough to separate two classes(a really harder question).

Even if I did not find any reference for this, I guess it is already well known that if C is closed under polynomes then \Sigma_j TIME(C)=TIME(C)^{\Sigma_j^P}, this is why I say that all of his discussion about the (weak?) exponential hierarchy is relevant to my question about alternating Turing Machine.

30. January 14, 2011 7:33 am

Most mathematicians are
too embarrassed to admit
that this very simple proof
of Beal’s Conjecture and
Fermat’s Last Theorem
is both true and correct.

http://donblazys.com/

I don’t blame them.

The sheer simplicity of the
actual solution to the worlds
most famous math problem
litterally stuns them and
makes them feel rather stupid!

Don.

• Christopher Heckman permalink
October 31, 2011 12:33 am

FLT has a long history of incorrect proofs, and non-proofs by amateurs. (The Four Color Theorem is another result with a similar history.) So when someone comes up to mathematicians with a FLT proof, the first reaction is, “Oh dear God, not another one.” So cut out your bad-mouthing.

What you need to ask yourself is: What, in your proof, would not have been known to, say, Gauss (or derivable by him)? If you can’t point to anything in particular, then your proof is probably wrong, because hundreds of people probably thought of a similar idea before and discarded it.

The consensus is that, unless some short proof of FLT (or the 4CT, for that matter) contains a genuinely new idea, it’s wrong. (I happen to share that opinion.)

One other question: Have you submitted your proof to a journal that referees its publications?

***

Getting back to the original topic, my nomination for an ME in Graph Theory is Steinberg’s conjecture: Any planar graph without cycles of length 4 or 5 is 3-colorable. (Grotzsch’s Theorem says that any triangle-free planar graph is 3-colorable, and it is obvious how the triangles in one of Steinberg’s graphs needs to be colored; there should be a simple extension of that coloring to the rest of the graph, but …)

A Steinberg Conjecture website that desperately needs to be updated is
http://math.asu.edu/~checkman/Steinberg.html , which lists results similar to the conjecture.

• November 29, 2011 12:22 am

Telling the truth is not “bad-mouthing”. Had you actually visited my website and read the proof, letters and articles, then you would have found that my proof does indeed contain a genuinely new idea called the “cohesive term”, and that it was indeed sent to the Journal of the London Mathematical Society whose referees supported it. That it was not published due to a “lack of journal space” only adds to the embarrassment!

The important thing is that my proof is exeedingly simple and both true and correct, which is why Google searching “Beal’s Conjecture Proof” shows that it is now ranked at the very top. Thus, the day is quickly approaching when it will become painfully clear to the math community that their so called “scholarly journals” are of little or no consequence because the vast majority of people actually use Google to look up new results in math.

Moreover, it is now becoming widely known that the Mathematics Department at Princeton University also knows about my proof, but is doing their very best to “sweep it under the rug”. Clearly, they can’t handle the embarrassment of having to tell their students that the algebra they are teaching is in actuality quite
inadequate for solving deeper problems such as Beal’s Conjecture.

You know, the internet is a very powerful tool, and with all the science and math forums out there, it would be very, very easy to set up an “online debate” between myself and the Mathematics Department at Princeton, or any college or university for that matter. That would be the fair and honest way to resolve, once and for all, the issue of my proof and the controversy it is engendering.

I am more than willing to participate in such a debate.

Unfortunately, every college and university math department is afraid to debate me, and that, in and of itself, is quite embarrassing, for it implies that they are all
nothing but cowardly little mice and scared little chickens.

Don.

November 29, 2011 1:25 am

Don Blaze wrote:

“Telling the truth is not ‘bad-mouthing’. Had you actually visited my website and read the proof, letters and articles, then you would have found that my proof does indeed contain a genuinely new idea called the ‘cohesive term’, and that it was indeed sent to the Journal of the London Mathematical Society whose referees supported it. That it was not published due to a ‘lack of journal space’ only adds to the embarrassment!”

That *does* sound suspicious, I’ll admit.

“The important thing is that my proof is exeedingly simple and both true and correct, which is why [...]”

Uh, that bad-mouthing thing, you did it again there.

• Bill Fahle permalink
January 17, 2012 11:51 am

What sounds suspicious is the handwaving in your proof where you assert without proof that (T(c/T))^(((3ln(c)/ln(T))-1)/((ln(c)/ln(T))-1))=c^3 has no solution. c is an integer>0, and T is any real, that is all we know for sure, so how do you reach this conclusion?

• March 3, 2012 1:53 am

Sorry Bill, but the only one doing any “handwaving” around here is you. The above equation is both WRONG and out of context. Apparently, you can’t even COPY without making mistakes! How embarrassing is that?

Don.

January 17, 2011 4:42 pm

Any knowledgeable mathematician would be embarrassed to claim a proof that consisted of:

1) Deriving an “identity” where the derivation is invalid at a specific value of T,
2) Substituting that “identity” into an equation ivolving T,
3) Discovering that the substituted equation can’t be evaluated at that specific value of T,
4) And then claiming that the specific value of T is “prohibited” in the original equation.

But that is exactly what you are doing, Don. It isn’t Beal’s equation that is invalid for the values in question, it is your so-called “identity.”

32. January 19, 2011 3:02 am

Kids have even a harder time with embarrassment than do adults. A lot of them desperately want to “hang out” on line with the “big boys”, but they know that sooner or later, they will make complete and utter fools of themselves, so in order to avoid being embarrassed, many of them developed the following rather transparent “strategy”.

They post using fake names like “Pipsi Cup”, “Foo Foo”, “Wumpy”, “Fingaboi”, “Mutt” and “Jeff Jo”. That way, they feel completely safe in spouting any kind of gibberish that comes to their young and inexperienced minds, without any possibility whatsoever of being embarrassed by it!

They are absolutely oblivious and entirely clueless as to what a cowardly act it really is to attack someones ideas from behind a wall of anonimity.

“Knowledgeable mathematicians” would never ever do that!

Note that R.J. Lipton, Terrence Tao and many others here are not afraid to use their real names, because they are confident in their abilities as mathematicians. Also note that no one here using their real name has dared to say anything even remotely negative about my proof.

That’s because these fine men are true mathematicians who can tell at a glance that my proof is both true and correct, and would never ever embarrass themselves or tarnish their names and reputations by attempting to refute something that they know is absolutely irrefutable.

These fine men are not cowards! They are not afraid to stand up for what’s right! They are not afraid to stand up in front of their peers and shout “Don Blazys is right! The properties of logarithms do indeed require that both Beal’s Conjecture and Fermat’s Last Theorem be true… and some long, drawn out, lawyer-like argument isn’t even necessary!”

Most importantly, these fine men know that to do or say nothing when the truth about the worlds most famous math problem is sitting right under their noses is perhaps the most insideous and lowly (and therefore embarrassing) form of cowardice that there is. Therefore I am quite certain that these fine men will verify and confirm my proof… right here, on this very blog… any day now!

Don.

• January 19, 2011 4:04 am

Don, I haven’t read, nor do I intend to read your possible proof, nor do I even care about Fermat’s Last Theorm (I’m a scientist, not a mathematician), but ad hominem attacks and grand procalamations of your own greatness will not encourage people to take your work seriously. Regardless of how great your work is, or how influential it could be, if you behave like a crackpot, people will assume that you are a crackpot. Please try to keep the discourse civil and grounded in reality.

• Alex Monras permalink
March 5, 2012 9:55 pm

In fact, he probably scores quite high in the crackpot index:
http://math.ucr.edu/home/baez/crackpot.html

• Carl Heckman permalink
March 8, 2012 1:22 am

(Well, actually responding to Alex Monras):

Nice! However,

(1) It would be nice to have a math version of the list; item #4 would count much heavier than 3 points.

(2) My favorite item is #12: “mailing your theory to someone you don’t know personally and asking them not to tell anyone else about it, for fear that your ideas will be stolen.”

Because, if you don’t want your ideas to be stolen, you definitely shouldn’t NOT tell everyone about them. (End sarcasm.)

33. January 19, 2011 8:23 am

Thanks Neil,

JeffJo got embarrassed a year ago when failed to refute my proof in another forum and has since become obsessed with my work. Thus, he now follows me around, all over cyberspace, still trying to refute it.

What would you do?

Don.

34. January 23, 2011 12:20 am

Come on Neil, this is not a difficult question, and you have now had several days to think about it. What would you do if you were being hounded throughout all of cyberspace by some kid calling himself “JeffJo”, who, having thoroughly embarrassed himself trying to refute an irrefutable result, is now simply resorting to ad hominem attacks, in some misguided and wildly irrational effort to soothe his bruised ego? Would you not make that attackers intentions and motives known to your readers, and thereby clear the way for a truly constructive, intelligent and civil dialouge? Well, that is exactly what I did, and as we all can see, “JeffJo”, (whoever he is) stuck his tail between his legs and scurried off!

Look, I know it seems like I’m being rather hard on poor “JeffJo”, but you see, in mathematics, truth is all we’ve got, and newly discovered truth, because it is still relatively unknown and in that sense “fragile”, must be defended vigorously and at all costs against those who would seek to obfuscate it, or otherwise prevent its dissemination. Moreover, in order to “survive infancy” and become established, newly discovered truth must also be promoted vigorously, even outlandishly, to those who are not quite bright enough to recognize it at first glance. It is only common sense that unless these things are done, mathematics will never advance or make any real progress.

Now, for the rest of us, let’s get back to the topic of “mathematical embarrassments”.

I would venture to guess that if we asked Grigory Perelman about what he considered to be the greatest mathematical embarrassment of all, then he would not hesitate in pointing to the math community itself! After all, by his actions (which speak far louder than words) he essentially told the entire math community to “stick their fields medal where the sun don’t shine” and to “take their funny money and go shopping”! Clearly, it was his way of pointing out that the professional mathematical community has far more than its share of pompous buffoons, and that he would rather be poor and live with his mother than to have anything whatsoever to do with them.

How embarrassing is that!

Now, multiply the intensity and scope of that embarrassment by a thousand and that’s pretty much what the math community will be facing when, sooner or later, they will have to explain to the general public, why the exeedingly simple, one page solution to the worlds most famous math problem was ignored for almost 12 years, even after it was refereed by the Journal of the London Mathematical Society, and found to contain no fatal flaw!

There will be no excuse, because my proof is in no way “obscure”. In fact, if you “Google search” the words “Beal’s Conjecture Proof” then you will find that my proof occupies the top three spots! Moreover, several versions of my proof, along with all the supporting documentation (including the letter from the London Mathematical Society) are right there on my website!

So Dick, Terrence and all the rest of you in the math community… go ahead and ignore this most simple result. (Your silence speaks volumes!) Just know that you do so at your own risk.

Don.

• January 23, 2011 5:33 pm

I just figured your question was rhetorical, since it makes sense as a rhetorical question: you’d already made your point. That and I’m more than too busy already.

I’m no stranger to being harassed, and I know how much the status quo of science has gone downhill (the same could be true of math as well), so I can sympathize with your frustration. If you can get other notable mathematicians on your side, it might help your case; how to do that, I have no idea. In terms of harassment, don’t feed the trolls… unless you’re also a troll.

However, I can guarantee that there is no conspiracy against you by “Dick, Terrence and all the rest … in the math community”. They already don’t have time to check the dozens of “proofs” that P=NP and the dozens of “proofs” that P!=NP.

That’s my last word on the subject.

35. January 25, 2011 7:46 am

I never suggested or even implied that there is some kind of “conspiracy by Dick, Terence and all the rest… in the math community.” I merely stated the easily verifiable fact that they have not yet responded, not even with a friendly “Hello Don”.

My guess is that they simply can’t respond, even if they wanted to, because the sheer simplicity of the proof, which can be found here:

http://donblazys.com/

makes the entire matter of Beal’s Conjecture andFermat’s Last Theorem far too embarrassing to even contemplate, much less discuss!

I honestly don’t blame them, because I understand perfectly well that this extraordinarily simple result, (which can be understood by any sufficiently bright high school student) really does put them, along with the entire math community squarely between “a rock and a hard place”.

Also, please be assured that I am neither a “troll” or a “crackpot”. Trolls don’t use their real names or make available their contact information and crackpots are simply wrong.

I invite any and all to go to my website and decide for themselves.

Don.

• January 28, 2011 6:45 am

Hi Don,

I took your invitation. In your note at the end (hereinafter N), which is part of your derivation of (5) and (6), we see that you have implicitly assumed T not equal to c, as otherwise the second part of (N) does not work; the denominator of the power would then be ln(c/c) = ln(1) = 0, and as I am sure you know division by zero is not very safe.

Since (N) is part of the derivation of both (5) and (6), the latter two results also hold only when T is not equal to c. Hence, the inability to substitute (c/c) for (T/T) in either (5) or (6) will naturally be a result of this property of (N), and does not imply that (c/c) and (T/T) are themselves inequal. Therefore no contradiction is reached, the paper just forgets to take into account the restriction of (N) in this case.

I hope that helps! You still have a framework for separating out the z=1 and z=2 cases, so maybe you can get something interesting out of it with some more work.

January 27, 2011 11:37 am

No, Don, you really don’t invite people to decide for themselves. You want them to say your proof is correct, and no one ever has (your claim that it was refereed is based on a politely worded rejection letter that said a referee had looked at it, not that it passed). When they point out any of the many and varied errors, you resort to ridiculing them. That is not very inviting behavior, and most people will not reply because they know it will happen. I just have a thick skin.

Or, in the only case of what you accuse me of above, you run to another site, misrepresent the issue, misinterpret the responses, and come back to claim some “expert” says you are right. Also not very inviting. I felt I had to post there, to defend myself and figure out what was actually said.

When I coincidentally tripped across this blog, googleing for threads involving proof by contradiction, I was surprised to see your comment there applauding rjlipton for his cautionary tale about the dangers involved. Quite ironic, since what he warned against is exactly what you do. I replied here only because it was more recent. I won’t any more, and I apologize to you all for doing it in the first place.

37. January 31, 2011 6:14 am

Thanks GK,

All sincere responses help in one way or another, and are greatly appreciated.

Now, in my view, if the inability to substitute (c/c) for (T/T) in either (5) or (6) does not imply that (c/c) and (T/T) are unequal, then what I “really” discovered is a way to prevent the substitution of 1 for 1, by a rather simple application of the properties of logarithms! That would make me a mathematical “miracle worker”!

Also, at the bottom of page 2, it says: “equation 7 is the case where z=1″ and “equation 8 is the case where z=2″. Thus, “seperating out the z=1 and z=2 cases” was done.

Don.

• February 2, 2011 6:14 am

Ah, I think I see where you are confused. It is partly a question of labels. As you know, the symbols we use to represent quantities in algebra are almost irrelevant; typically we use “x”, “y”, and so on, but we could equally use any other symbol, or even a picture of a flower if we wanted. It is just a label. The problem you have run into here is that the choice of label is only “mostly” irrelevant. I will give an example.

Suppose I have a function f(x) = 2x. My choice of the label “x” is of course arbitrary. Now if I integrate this with respect to x, I receive x^2 plus a constant. My choice of label for the constant is only MOSTLY arbitrary, because there is one label I cannot use without confusing things: I can’t very well call the constant “x”! This is because “x” is a /bound variable/; it already has a defined meaning and context within my current calculations, and therefore I must stick to this context or else I will not have a consistent system and therefore can’t expect consistent results. The integral of f(x)=2x is, as you know, certainly not (x^2+x) by any means!

In your paper, the label “c” has been initially bound during (1). T is bound during (4). So far there is no real conflict here. However, in the first part of (N) you use a property of logarithms which only holds if c and T are not equal; it is not a totally universal identity. This restriction is now part of the context of c and T; further equations must stick with this context, if they don’t then a contradiction may arise (as did happen). So, although the choice of label for “T” is mostly arbitrary, in (N) you have essentially stated that the label “c” cannot be used for it; “c” is already in use and its context must be obeyed!

Note that the restriction (c not equal T) is a result of the property of logarithms! It applies any time you use this property, it is not just related to Beal’s conjecture. This is an unfortunate problem with reductio ad absurdum; if there is more than one source of contradiction you can’t tell which one causes the absurdum! Don’t worry, lots of people have run into this; it can sometimes be hard to spot “hidden” restrictions like the one on this logarithm property.

Keep trying! Most people do not put in as much thought in mathematics as you have; you can even be proud of this mistake because most people will never think about things hard enough to even run into it.

• February 2, 2011 6:27 am

I wonder if schools should give more time to consideration of such matters. In complicated algebra, keeping track of context can be quite important. Consider the following short “proof” (I am sure you will all spot the mistake immediately, by watching what is happening in the context of the restrictions placed on “x”).

:: x = 5

Multiply by x:
=> x^2 = 5x

Subtract 25:
=> x^2 – 25 = 5x – 25

Factorise:
=> (x-5)(x+5) = 5(x-5)

Cancel out (x-5) term:
=> (x+5) = 5

Substitute value of x:
=> 10 = 5 ???

Always be careful with division operators, they are dangerous weapons.

38. February 3, 2011 7:36 am

Thanks GK, for all your hard work, time, patience and wonderful conversation.

Now, in your commentary, you keep referring to the “derivation” which, for the sake of convenience, we are calling (N). Please understand that (N) is only a “note”, uses only one case as an “example”, and is not part of the proof. The key properties of logarithms involved in (N) could just as well be applied simultaneously, as they are in the proof itself.

In your second paragraph, you wrote, “…in the first part of (N), you use a property of logarithms which only holds if c and T are not equal”.

Please note that, in the actual proof, we can let T equal c if and only if z=1 or z=2.

Right after that you wrote, “…it is not a universal identity”.

Please note that the identity is true for all T=2,3,4…, c=1,2,3…, and z=1,2,3…, and is therefore sufficient to encompasses all possible cases of Beal’s conjecture, which specifies positive integer variables only.

The next sentence in your commentary, “This restriction is now part of the context of c and T; further equations must stick with this context, if they don’t, then a contradiction may arise (as did happen)”.

Again, the restriction T not equal to c only applies if z>2. If z=1 or z=2, then that restriction does not exist (which proves the conjecture).

The beginning of your next paragraph, “Note that the restriction (c not equal T) is a result of the property of logarithms! It applies any time you use this property, it is not just related to Beal’s conjecture.”

Again, the restriction (c not equal T) exists if and only if z>2, but here I must also respectfully disagree with you because the properties of logarithms are simply a logical extention of exponentiation and must therefore be consistent with the properties of positive integers. In other words, if a counter example of Beal’s conjecture did exist so that x,y,z > 2, then logarithms would have to be eliminated from mathematics entirely because their properties would not allow that!

Even though we disagree, I thank you with all my heart for your commentary.

By the way, that short proof involving ::x=5 was absolutely delightful. I explained it to my grand daughter who is just beginning to study algebra and she absolutely loved it!

Don.

August 7, 2011 7:23 pm

Don’s proof is seriously flawed. Many competent mathematicians have tried to point this out to him, but he fails to listen. To save everyone the time, Don’s proof assumes that 1^(0/0) is both defined and determinate, which is not true.

• November 29, 2011 3:01 am

How tiresome.

Yet another anonymous and therefore cowardly “pot shot” at my proof!

Are you by any chance “Marky Mark”, or are you perhaps the “Mark of Zorro”?

Anyway, it is poor “Mark” (whoever he is) who is clearly wrong.

As anyone can see, my proof, which can be found here:

http://donblazys.com/

does not even contain the indeterminate form 0/0. Apparently, “Mark” never even learned how to simplify expressions so as to avoid such pesky indeterminate forms.

Moreover, “Mark” fails to realize that if n*0 = 0, then 0/0 = n is a true statement for any n, which means that 1^(0/0) = 1^n = 1 is indeed both defined and determinate!

Don.

December 23, 2011 12:01 pm

Leave it to Don to insult people who find flaws in his proof.

If n can be any value, then n can be infinity. 1^infinity is indeterminate.

• Proginoskes a.k.a. Christopher Heckman permalink
December 29, 2011 3:43 am

Actually, that’s not the only flaw in his proof. I’m surprised that no one has mentioned that

(1) He has artificially set up his formulas to make them work for z=1 and z=2 but nothing else. (Note, it’s also possible to set up the formulas to make them work for z=3, z=4, or z=any integer greater than 1), or that

(2) In his proof, it doesn’t really matter what c^z is equal to (so his method proves that there are no solutions to c^z = 27 with z>2, either, which implies that his method is invalid).

— Christopher Carl Heckman

• November 29, 2011 1:28 am

Please note that “GK”, as nice as he is, was also afraid to “refute” my proof using his real name.

That’s because “GK”, (whoever he is), couldn’t really find a fatal flaw, so he merely posted some pot shots and gibberish under the pseudonym “GK” (short for “GEEK”?), knowing full well that such pot shots and gibberish would embarrass him greatly and forever tarnish his reputation if he used his real name.

Well, maybe we don’t know who he is, but he most certainly does, and he must now come to terms with his own cowardice.

Don

39. Dan L. permalink
January 4, 2012 11:53 pm

I think Don has invested so much pride into his result that he is afraid and embarrassed to admit that there might be an error. This is especially true when he has an entirely friendly back-and-forth with GK and then turns around and seems to get really paranoid, assuming GK is pseudonymous out of fear and insisting that GK was responding with “pot shots and gibberish.” This is also clear from the way he fiercely tries to shout down any opposition.

GK’s objection might not be valid, but it is certainly not gibberish. I’m not convinced it defeats the proof but it looks to me as though it should. But I’m not an experienced, published mathematician so I’m entirely willing to admit I might be wrong. I do know that GK’s objection was not a pot shot (GK was very precise about what the problem might be) and it was not gibberish (I think I understand it and I think it’s correct). But again, not superman so I could be wrong. I’d leave it to the experts.

What about you, Don? Can you admit even that you MIGHT be wrong? Can you leave it up to the experts to decide?

• March 2, 2012 5:25 am

Of course I can admit when I am wrong! After all, I am using my real name and would not tarnish it by supporting a proof that was fatally flawed. Trust me, if someone actually did find a fatal flaw in my proof, then I would immediately apologize to everyone, take it off my website and have it removed from the online journal “Unsolved Problems”. In other words, I would drop it like a hot potato!

However, the FACT remains that after more than a dozen years, no one has ever come up with a reasonable or valid objection to my proof. GK’s objection is clearly invalid, as are Christopher Heckman’s and “Mark’s” (who doesn’t even understand that infinity is not a number!).

In fact, Google searching “Beal’s conjecture proof” shows that all three versions of my proof are now ranked at the VERY TOP! This is a big “slap in the face” to my critics and a downright embarrassment to the entire “math community”, especially since most students go to Google (and not some obscure “scholarly journal”) to learn about Beal’s conjecture.

Thus, it is IMPORTANT that this issue be settled once and for all, and that the truth be known. So… here’s what I propose…

I will DEBATE the merits of my proof with the mathematics department of any major college or university. All I ask is that it be in a public forum with a fair and impartial moderator, and that my opponents use their real names.

Surely, someone here can contact some well known mathematicians such as Terence Tao or Andrew Wiles and inform them of my challenge to debate not only them, but any “team of renowned mathematicians” that they might want to put together!

Besides, think about how much fun it would be!

Don.

• Carl Heckman permalink
March 3, 2012 3:10 am

“No one has ever come up with a reasonable or valid objection to my proof. GK’s objection is clearly invalid, as are Christopher Heckman’s.”

Hey, YOU stopped our email exchange. And I have not one but THREE arguments, none of which you have refuted.

IOW, your claim is only true because you’ve put your fingers in your ears and closed your eyes.

— Christopher Carl Heckman

(posting with my facebook account)

40. March 4, 2012 12:55 am

To Christopher Carl Heckman,

Here’s a true story.

Like me, Albert Einstein had a lot of critics. A group of them got together and published a book entitled “100 Authors Against Einstein”. When asked about what he thought of that book, Einstein replied “If I were wrong, then it would only have taken one”.

Now, you claim to have found THREE arguments against my proof! Well, like good old Albert said, all it would take is one!

However, as I already explained to you during our e-mail exchanges, NONE of your arguments have any merit whatsoever. “Indeterminate forms” are NOT an issue and you CAN’T simply ignore the REQUIRED second degree radical, so it’s NOT possible to “set up the formulas to make them work for z=3, z=4 etc.” as you assert.

As for ending our e-mail exchanges, please understand that ever since Google ranked all three versions of my proof at the VERY TOP, I have been receiving more e-mails than I, a “one finger typist”, can possibly answer. All kinds of e-mails, from curious students, nasty critics and kind supporters, to cranks who want me to look at their “work” and women who want marry me have been pouring in, and I simply don’t have the time to read them all, much less answer them.

But… if you are serious, and you still think that you have a valid argument, then I WILL make time for an online debate at some open and public forum with LaTex capability. We can then invite a large portion of the math community, along with members of the media, and thereby settle this issue once and for all. Just pick the ONE (and only one) argument that you believe to be absolutely valid, and if I can’t counter it with logic, then you will win and I will renounce my proof! My only advice to you would be to pick a TEAM of really good mathematicians to help you, because even the EXPERT number theorists at the London Mathematical Society were unable to find a “fatal flaw” and were thus forced to give my proof some support!

Also, there is now this to consider. By this time next year, this whole saga might be over anyway. You see, one of the more interesting e-mails that I recieved recently was from a college student who thinks he can verify my proof using an “automated proof checking program”. I met with him about a month ago, and we both decided that it would be a lot of fun to see if could actually be done.

Don.

• Carl Heckman permalink
March 4, 2012 2:47 am

Oh my … oh my … where to start …

“However, as I already explained to you during our e-mail exchanges, NONE of your arguments have any merit whatsoever. ‘Indeterminate forms’ are NOT an issue”

Wrong.

“and you CAN’T simply ignore the REQUIRED second degree radical, so it’s NOT possible to ‘set up the formulas to make them work for z=3, z=4 etc.’ as you assert.”

Wrong. I showed you how to do this in my email. All you have to do is replace ’2′ with ’3′ (or some other number) in your formulas.

You’re forgetting the third mistake as well.

“As for ending our e-mail exchanges, please understand that ever since Google ranked all three versions of my proof at the VERY TOP, …”

That’s your own fault. Or your webmaster’s. Google rankings don’t happen by accident.

“But… if you are serious, and you still think that you have a valid argument, then I WILL make time for an online debate at some open and public forum with LaTex capability. We can then invite a large portion of the math community, along with members of the media, and thereby settle this issue once and for all.”

I don’t know about the media — they don’t tend to care about things like this this. But if you want to to discuss this in a ‘public form with LaTeX capability’, then I suggest http://forums.xkcd.com/ . (This is run by the guy who writes the comic; there are many science-related subforums there as well. It is a highly mathematically-literate community. They use jsMath in their posts.)

“Just pick the ONE (and only one) argument that you believe to be absolutely valid, and if I can’t counter it with logic, then you will win and I will renounce my proof!”

I will not say any more about the “indeterminate form” flaw, since there are some sticky issues involved there. However, I do want to use my two other arguments. The more powerful one — that your proof really proves a lot more than what you say it does — may be over your head.

And now for my conditions: You have to restrict yourself to standard logic. I reserve the right to indicate when you are not using proper logic, and add that if you are using logic incorrectly, then you must admit defeat and remove your website.

“You see, one of the more interesting e-mails that I recieved recently was from a college student who thinks he can verify my proof using an ‘automated proof checking program’. I met with him about a month ago, and we both decided that it would be a lot of fun to see if could actually be done.”

I’d also like to see the results.

See you at forums.xkcd.com.

— Christopher Carl Heckman

• March 6, 2012 3:31 pm

To Christopher “Carl” Heckman,

Of course you will “not say any more” about the”indeterminate form flaw”. That’s because there is no “indeterminate form flaw”! The fact is, in my proof, if the equations are properly simplified, then “indeterminate forms” DON’T EVEN EXIST, and that, as any schoolboy knows, makes them an UTTERLY TRIVIAL NON-ISSUE. Thus, if you did “say any more” about them, then you would appear even sillier than you already do!

As for your “conditions”, well, this is to be a well publicized FORMAL DEBATE, so we will have a panel of EXPERTS, all using their real names,
indicate if and when proper logic is not being used. That’s only fair.

Also, since this debate will be about my TOP RANKED PROOF of Beal’s Conjecture, (which automatically proves Fermat’s Last Theorem) I’m quite sure that a lot of science writers will indeed find it interesting, and at the very least, monitor the proceedings. Thus, I think we should vigorously PROMOTE the event and invite every college and university math department, every mathematical magazine, every math and science forum, every math blog, etc.

Don.

41. Carl Heckman permalink
March 7, 2012 2:41 am

Don Blazys left a reply, but didn’t allow me to reply to IT. So here we go:

“To Christopher ‘Carl’ Heckman,”

Don’t put Carl in quotation marks; that’s my real, legal, middle name. Also, there is a Christopher A. Heckman who also has a degree in mathematics, so I include the ‘Carl’ to avoid confusing people. (The NSA certainly got confused by the two of us.) Check out my website at http://math.asu.edu/~checkman/ .

“Of course you will ‘not say any more’ about the ‘indeterminate form flaw’. That’s because there is no ‘indeterminate form flaw’! …”

Yeah, yeah, yeah. Save it for the debate.

“As for your ‘conditions’, well, this is to be a well publicized FORMAL DEBATE, so we will have a panel of EXPERTS, all using their real names,”

No real experts would spend more than 15 minutes on your paper, for reasons discussed elsewhere on this page. The community at forums.xkcd.com has already had a discussion about Beal’s Conjecture, incidentally.

“Also, since this debate will be about my TOP RANKED PROOF”

Yeah, yeah, we know how good your webmaster is at using meta-tags.

“… of Beal’s Conjecture, (which automatically proves Fermat’s Last Theorem) I’m quite sure that a lot of science writers will indeed find it interesting, and at the very least, monitor the proceedings. Thus, I think we should vigorously PROMOTE the event and invite every college and university math department, every mathematical magazine, every math and science forum, every math blog, etc.”

You go ahead and do that; you’re evidently the expert at promotion. I’m busy the next few weeks, teaching at Arizona State University.

BTW, who was the student who had the automated proof checking program?

• May 23, 2012 2:20 am

To: Christopher Carl Proginoskes Heckman,

In this topic alone, you have gone by the names Carl Heckman, Christopher Heckman, Christopher Carl Heckman and Proginoskes. Heck man, even most “gansta rappers” and pole dancers don’t have that many names! It’s no wonder that the NSA got confused with regards to your identity!

Anyway, whoever you are, the truth is, I don’t have a “webmaster”, and I don’t even know or care to know what “meta-tags” are. Google ranks my proof at the very top simply because it happens to be both true and correct… and people respond to that!

I am by no means an “expert on promotion” and my proof dominates because it is a genuine breakthrough.

So you see Christopher Carl Proginoskes Heckman, unless you are able to set up a debate that suits my fancy, and involves not only the media but at least several “famous” mathematicians like Terence Tao or Andrew Wiles, I will just continue to ignore you.

After all, the longer I wait while enjoying my top rankings on Google, the more embarrassing it is for the rest of you!

Doesn’t that make you happy?

Don.

• Carl Heckman permalink
May 23, 2012 8:44 pm

Skipping over the insults … Skipping over the “I don’t have a webmaster” irrelevancies … Skipping over the “expert on promotion” comments …

Are you finally ready to discuss your proof, or are you going to continue with ad hominems? Because YOU were the one who was going to choose the panel of mathematicians to judge the outcome — we agreed on this.

And acting like this makes it look like you’re scared of what’s going to happen, your “proof” going down in about five minutes.

I haven’t had a single thought about “Don Blazys” the past two months.

When you’re ready to discuss real math, let me know. Stop talking smack, like you’re doing here.

42. May 25, 2012 5:01 am

To: Christopher Carl Proginoskes Heckman,

It is you who is “talking smack”.

The truth is this.

You are a coward because you attack other peoples ideas from behind a veil of anonymity, using phoney and childish names like “Proginoskes”.

You are also a pompous buffoon who not only “steps in it” at every turn, but then puts that same foot in his mouth. You say venomous things like “No real experts would spend 15 minutes on your paper”, when YOU yourself already spent SEVERAL MONTHS debating my proof via e-mail! What kind of convoluted logic is that?

Your eager willingness to “continue” our debate can ONLY be explained by the fact that YOU KNOW YOU LOST. In fact, your arguments were so pathetic that I actually felt sorry for you!

Any real mathematician can see AT A GLANCE that my proof is both true and correct. Thus, no real mathematician will ever debate me because only a fool would argue against a result which is so obviously both true and correct!

Since you are the fool itching to continue our debate, it should be you who does most of the work publicising it. Besides, as a “professional” who works at a “major university”, it is you who has the means to contact “famous” mathematicians such as Terence Tao.

For my part, I will try contacting him too… right now.

Hey Terence Tao, where are you?

Can you tell this clown that my proof is indeed both true and correct?

Come on Terence, any coward can ignore the truth.

Let’s see if YOU have the COURAGE to stand up for it!

Don.

• Christopher Carl Heckman permalink
May 25, 2012 6:53 am

This nonsense has gone on long enough.

Any legitimate proof of Beal’s Conjecture must use the co-primality in an essential way; there are solutions if co-primality is not assumed (3^3 + 6^3 = 3^5, for instance).

The only place where Blazys “uses” co-primality is in the equation 1 = T/T, which in fact does not rely on that assumption in order to be true.

Therefore Blazys does not use co-primality in an essential way.

Therefore Blazy’s proof is incorrect.

QED.

(Oh, and “Don”, I’m sure that your real name, on your birth certificate, is something like “Donald”.)

As for the previous post:

“You are also a pompous buffoon who not only “steps in it” at every turn, but then puts that same foot in his mouth. You say venomous things like “No real experts would spend 15 minutes on your paper”, when YOU yourself already spent SEVERAL MONTHS debating my proof via e-mail! What kind of convoluted logic is that?”

I know that your proof is wrong. I am using my time to try to show you why your proof is wrong.

As for: “Since you are the fool itching to continue our debate”, well I didn’t have a single thought about your unsuccessful attempt at immortality, until you posted here. So in fact, you are the one who wants to continue the debate.

Your past three posts have included no mathematical content. Do not post again until they do.

• May 26, 2012 6:44 am

To: Christopher Carl Proginoskes Heckman,

I am both honest and consistent in all things, including my identity. My first name is not Donald or Donovan. It’s Don.

You, on the other hand, are a liar, a sock puppet, and the fool itching to continue our debate.

The fact is, you began e-mailing me long before I ever posted here.

Anyone can simply scroll up on this very topic to March 3rd 2012, and see for themselves where you wrote, ” Hey, YOU stopped our email exchange”.

Thus, even as a lying sock puppet, you are utterly incompetent and unable to fool anyone!

As for your claim that my proof “does not use co-primality in an essential way”, well, that just demonstrates that you are a mathematical bumpkin.

My proof of Beal’s Conjecture, which can be found here:

http://donblazys.com/02.pdf

clearly states, “let a, b, c be co-prime, so that the only common factor possible is 1 = T/T”.

Thus, everyone but you can see that in my proof, co-primality is indeed both assumed and used in an essential way. The absence of any non-trivial factors is both explicit and implicit.

Solving a math problem does not make one “immortal”, but it does make one popular enough to be consistently ranked at the very top by Google!

Doesn’t that make you happy?

Don.

43. Christopher Carl Heckman permalink
May 27, 2012 10:28 am

Don Blazys wrote: “My proof of Beal’s Conjecture, which can be found here:

http://donblazys.com/02.pdf

clearly states, “let a, b, c be co-prime, so that the only common factor possible is 1 = T/T”.

Thus, everyone but you can see that in my proof, co-primality is indeed both assumed and used in an essential way.”

That’s not an essential use, though, and you missed the point entirely. (This is one reason why people stop talking to you, Don.) You *could* eliminate the phrase “let a, b, c be co-prime, so that the only common factor possible is”, since 1 = T/T anyway, *regardless* of what gcd(a,b,c) is. That fact that you felt it necessary to put it in does not change the fact that co-primality is nonessential to your proof.

And Google has evidently decided that your proof is wrong, because you’re not listed “at the very top” any more.

On a more serious note, what do you think of James Constant’s Proof, at http://www.coolissues.com/mathematics/Beal/beal.htm ? He beat you to the goal, man.

• Christopher Carl Heckman permalink
May 27, 2012 10:35 am

Additional: Don wrote: “The fact is, you began e-mailing me long before I ever posted here. ”

Lie. I never ever heard of Don before he posted here.

‘Anyone can simply scroll up on this very topic to March 3rd 2012, and see for themselves where you wrote, ” Hey, YOU stopped our email exchange”.’

Yes, when I was starting to convince you that you were wrong.

And Don, you need to get your insults right; look up the definition of a sock puppet, won’t you?

Maybe I’ll call up Terrence Tao and ask him … Uh oh, I don’t have his number. Gee, how can I contact him? I guess I’ll Google “Terrence Tao math email” and …

Gee! First link (literally, at UCLA)! There it is!

What’s the matter Don, couldn’t you figure that out yourself?

• Christopher Carl Heckman permalink
May 28, 2012 12:23 am

Blazy’s argument falls apart for a more fundamental reason. He proves that if a,b,c are coprime, and a,b,c,x,y,z are integers with
a^x + b^y = c^z,
then it is z that must be at most 2. However,

(1) 1^2 + 7^1 = 2^3
(2) 1, 7, and 2 are coprime, and
(3) 3 is not equal to 1 or 2.

So the argument is wrong by a counterexample.

• May 30, 2012 12:34 am

Here again, is my proof:

http://donblazys.com/02.pdf

Everyone exept Christopher Carl “PROGINOSKES” Heckman can see that assuming the conjecture false requires x, y, z all greater than 2.

Thus, not only is he wrong again, but he is now so utterly obsessed with my proof that he is “triple posting”!

Don.

• Christopher Carl Heckman permalink
May 30, 2012 10:03 am

Don is right about something: Assuming the conjecture is false assumes that x, y, and z are all greater than 2. He has finally proven his proof wrong, because his proof doesn’t assume this, only that z > 2.

According to his proof, I can’t do the calculation 2^3 = 8, because 3 can’t be an exponent, ever, in any expression like this.

Blazys has proven that 8 is not equal to 8, a feat which is hardly new and hardly correct.

And I am not obsessed; if Blazys had been smart enough to understand that his proof was incorrect — and why — I wouldn’t have to keep clarifying these basic issues. (I wonder if Blazys debates what the doctor says when he makes a medical appointment? “Mr. Blazys, you have a broken arm.” “No, it’s just an air bubble.”)

• Christopher Carl Heckman permalink
May 30, 2012 6:14 pm

Something else Don is right about: I simply have to stop responding to trolls.

44. May 31, 2012 2:01 am

Here, yet again, is my proof.

http://donblazys.com/02.pdf

It’s easy to see that when the conjecture is assumed false, x,y,Z and z are all greater than 2.

Thus, Christopher Carl “PROGINOSKES” Heckman is wrong yet again.

Only idiots respond to trolls.

Don.

45. James Tomson permalink
January 1, 2013 11:08 pm

For proofs of FLT and disproof of Beal’s conjecture see
Beal Fermat and Pythagora’s Triplets http://www.coolissues.com/mathematics/BealFermatPythagorasTriplets.htm

• Christopher Carl Heckman permalink
January 3, 2013 3:26 am

****, I wasn’t going to do this again, but …

Your proof falls apart because 27^4 + 162^3 = 9^7, 27, 162, 9, 4, 3, and 7 are positive integers, and 4, 3, 7 > 2.

— Christopher Carl Heckman