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Remembering that a STEP mathematics question which guided us through showing $n^3 + (n-3)^3 = (n+3)^3$ has no solutions in the integers could be sledgehammered with Fermat's Last Theorem, I wondered if there are other fun ways to apply Fermat's Last Theorem.

Another example:

Claim $ \sqrt[n]{2}$ is irrational for all integers n > 3

Proof Suppose to the contrary, that $ \sqrt[n]{2} = \frac{p}{q}$ for integers p and q. Then $q^n + q^n = p^n $. A contradiction (Wiles, 95).

Any other examples? I'd be happy to see applications of other big theorems (such as the Catalan conjecture/Mihăilescu's theorem). Fermat fits naturally because of the gulf in the elementariness of the statement and the difficulty of the proof.

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More generally $\sqrt[n]{a^n+b^n}$ is irrational if $a,b>0$, otherwise $p^n=(qa)^n + (qb)^n$. – Thomas Andrews Apr 12 '12 at 15:06
I'm sure there are examples that use the classification of finite simple groups. – Thomas Andrews Apr 12 '12 at 15:09
See a comment on circularity of this proof:… – sdcvvc Apr 12 '12 at 15:25

This so called application of Fermat's great theorem was given at a Romanian olympiad. It is not suitable for such a purpose, but I'm not sure why this was accepted as a contest problem.

Prove that if $n$ is odd, $a,b,c$ are non-zero integers and $$ a^{3n}+b^{3n}+3(abc)^n=c^{3n} $$ then $P(a,b,c)$.

Where $P(a,b,c)$ is a statement about $a,b,c$ which can be deduced from $a=b=-c$.

Proof: $$ (a^n)^3+(b^n)^3+(-c^n)^3-3(a^n\cdot b^n\cdot (-c^n))=0$$

and then we use $$ x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$$

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Beni Bogosel, where is FLT used? Is it somewhere from a = b = -c to P(a,b,c)? – BCLC Dec 10 '15 at 15:21
One of the factors given by the equation is $a^n+b^n-c^n$ which by FLT is never zero. – Beni Bogosel Dec 10 '15 at 18:07
So how to deduce $a=b=-c$? – BCLC Dec 10 '15 at 18:09
From the other part of the product $(a^n-b^n)^2+(b^n+c^n)^2+(a^n+c^n)^2 = 0$. – Beni Bogosel Dec 10 '15 at 18:15

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