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Problem

Let $x$ and $y$ be positive whole numbers, and let $p$ be any odd prime.

It is well known that $x^3 + y^3$ is never equal to an odd prime.

But given that $n$ is a positive integer which contains an odd factor greater than one, prove that $x^n +y^n = p$ has no solutions.

Solution

Let $V = x^n + y^n$.

As V is a prime greater than two it is clear that $x \ne y$, otherwise the sum would be $2x^n$ which is divisible by 2. In addition $x \gt 1$, in which case $x^n \gt x$.

So without loss of generality let $x \gt y \ge 1$. Therefore $1 \lt x + y \lt V$.

It can be verified for odd values of $n$ that:

$V = (x + y)(x^{n-1} - x^{n-2}y + ... + x^2 y^{n-3} - x y^{n-2} + y^{n-1})$

In other words, $V$ is divisible by $x + y$, which lies between 1 and $V$. Hence for odd values of $n$, $V$ cannot be prime.

Suppose that $n = ab$, where $a = 2^k$ and $b$ is an odd greater than one.

$\begin{align}\therefore V &= x^n + y^n\\&= (x^a)^b + (y^a)^b\\&= (x^a + y^a)((x^a)^{b-1} - (x^a)^{b-2}(y^a) + ... - (x^a)(y^a)^{b-2} + (y^a)^{b-1})\end{align}$

In the same way as before, $1 \lt x^a + y^a \lt V$, and as $V$ is divisible by $x^a + y^a$ it cannot be prime.

Hence we prove that $x^n + y^n$ is never equal to an odd prime unless $n$ is of the form $2^k$.

Given that $n = 2^k$, investigate when $V$ is prime.