Characterization of primes dividing $(x+y)(x+z)(y+z)$ in a [hypothetical] counterexample to Fermat's Last Theorem

Question

Assume $p$ is an odd prime, and $x,y,z$ are pairwise relatively prime nonzero integers, such that $x^p+y^p+z^p=0$.

In Ribenboim's Fermat's Last Theorem for Amateurs, he gives a proof (p. 101) that every prime divisor of $(x^p+y^p)/(x+y)$ is congruent to $1\!\!\pmod{\!2p^2}$; by symmetry the same is true also for $(x^p+z^p)/(x+z)$ and $(y^p+z^p)/(y+z)$.

QUESTION: Are there similar results about the size or shape of primes dividing $(x+y)(x+z)(y+z)$?

In particular, I'm interested in the following conjecture.

CONJECTURE: At least one of $x+y$, $x+z$, or $y+z$ must have a prime factor greater than $p$.

Answer 1

Here's an answer:

Assume $x^p+y^p+z^p=0$ for some integers $x,y,z$ and odd prime $p$. Then by Fermat's Last Theorem, we have a logical contradiction. It therefore follows that $1+1=2$, and also that $1+1=3$.

And also that $(x+y)(x+z)(y+z)$ is divisible by $47$. And that all divisors of $(x+y)(x+z)(y+z)$ are congruent to $1$ mod the order of the Monster group.

These are just some examples of divisibility criteria on $(x+y)(x+z)(y+z)$ that follow from the assumption.

Answer 2

Mod $2$, the assumption (that FLT is false) implies $x+y+z\equiv0$. Since $x,y,z$ are relatively prime, they can't all be even. So precisely two of them are odd. So at least one of the three factors in $(x+y)(y+z)(x+z)$ is even. Therefore, $2$ is a prime dividing $(x+y)(y+z)(x+z)$.

(But then if we remember FLT is true, we can also say $2$ does not divide $(x+y)(y+z)(x+z)$, because there are no triples satisying $x^p+y^p+z^p=0$ in the first place.)

Answer 3

In the first case of Fermat's Theorem one can apply Sophie Germain's method to obtain factors. For example, if $2p+1$ is prime then $2p+1$ must be a factor of $(x+y)(y+z)(z+x)$. Similarly, one can use other primes of the form $np+1$.