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An Old But Cool Result

February 6, 2019

Solving a type of Fermat Equation

Leo Moser was a mathematician who worked on a very varied set of problems. He for example raised a question about “worms,” and invented a notation for huge numbers.

Today I want to talk about one of his results with a very short proof.

No, it is not about worms. That is a question in discrete geometry that is still open I believe: “What is the region of smallest area which can accommodate every planar arc of length one?” The region must be able to hold the arc inside but the curve can be moved and rotated to allow it to fit. A disk of diameter {1} works and has area about { 0.78}. It is possible to do much better and get around { 0.27}.

See this paper for some additional details.

No, it is not about a conjecture of Paul Erdős See this for a great paper on this result:

Theorem 1 Suppose that

\displaystyle  1^{k} + 2^{k} + \cdots + (m-1)^{k} = m^{k}.

Then any {(m,k)} solution in integers with {k \ge 2} must have

\displaystyle  m > 10^{10^{6}}.

Erdős conjectured there are no solutions at all. It is easy to check that for {k=1} the unique solution is a bit smaller:

\displaystyle  1 + 2 = 3.

The Result

Yes, it is about the solution to a natural family of Diophantine equations. This result of Moser comes from an old paper of his. The result can be found on the wonderful blog called cut-the-knot written by Alexander Bogomolny.

The question considered by Moser is simple to state:

Consider the equation over the integers {x^{a} + y^{b} = z^{c}} where {a, b, c} are fixed values that are relatively prime. Show that there are infinitely many integer solutions.

The surprise, to me, is that this equation always has integer solutions. I thought about it for a bit and had no idea how to even start.

The solution is as follows. The initial insight is that the restriction on the exponents implies that there are integers {m, n} so that {abm + 1 = cn}.

Wait a minute. We must be careful by what we mean by “the values of {a,b,c} are relatively prime.” We need more than the greatest common divisor (GCD) of {a,b,c} is {1}. We need that {ab} and {c} are relatively prime. Note that {6,10,15} have GCD equal to {1} but no matter which of the triple is “{c}” we cannot find the needed {m,n}:

\displaystyle  (6\cdot 10,15) > 1, (10\cdot 15, 6)>1, (15,6 \cdot 10) > 1.

I thank Subrahmanyam Kalyanasundaram for catching this.

The next idea is not to look for a single set of solutions but rather to find a parametrized solution. That is try to find expressions for {x,y,z} that depend on some variables {u,v} so that for all {u,v} the equation is satisfied.

Then set

\displaystyle  x = u^{bm}(u^{abm} + v^{abm})^{bm}.

\displaystyle  y = v^{am}(u^{abm} + v^{abm})^{am}.

Note as {u,v} vary over integers the values of {x} and {y} vary over integers too. The claim is that this is a parameterization of the equation. Let’s see why. We need to figure out what {x^{a} + y^{b}} is equal to. It looks a bit nasty but it is not. Let {W = u^{abm} + v^{abm}}. Then

\displaystyle  x = u^{bm}W^{bm}.

\displaystyle  y = v^{am}W^{am}.

So {x^{a} + y^{b} } is

\displaystyle  u^{abm} W^{abm} + v^{abm}W^{abm}.

Which magically is

\displaystyle  W^{abm}(u^{abm} + v^{abm}) = W^{abm+1}.

Thus setting

\displaystyle  z = W^{n}

implies that

\displaystyle  x^{a} + y^{b} = W^{abm+1} = W^{cn} = (W^{n})^{c} = z^{c}.

Very neat. By the way we do need to note that as {u} and {v} run through integers the values of {x} and {y} and {z} vary enough to get an infinite number of solutions. A simple growth argument shows that this is true.

The key trick was to not use a standard idea and apply the binomial theorem and expand

\displaystyle  (u^{abm} + v^{abm})^{bm}.

My algebra DNA suggests that expanding such an expression is often a good idea. Here it would lead to a mess. This is a case where using the binomial expansion does not work.

Open Problems

I really like Moser’s clever solution to the diophantine equation

\displaystyle  x^{a} + y^{b} = z^{c}.

Note that it must fail when {a=b=c=p} for {p>2} by the famous solution to the original Fermat equation.

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