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Let $z,n,x,y,m$ be positive integers with $z \ge 5$ and $m \ge 3$ and $m$ odd.

Does it follow that:

  • $z$ cannot be prime

  • if $p \ge 5$ and $p | z$, then either $p > m$ or $p|m$

Here is my reasoning:

  1. If $z$ is prime, then there are no solutions as shown here.

  2. If $p | z$, $p$ doesn't divide $x$ or $y$ since if we assume $n$ is the lowest positive integer solution, we get a contradiction by the logic in step 1 if $p | xy$

  3. If $p \le m$, then using Lifting the Exponent (see Theorem 2), $p$ must divide both $x+y$ and $\dfrac{x^m + y^m}{x+y}$ and there is no solution unless $p | m$.

Am I correct? Is there a mistake in my reasoning? Is there a counter example?

Edit: added details to explain why LTE applies based on comments received.

Edit 2: for this logic to be correct, each $p \ge 5$.

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Larry Freeman
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  • You can only use LTE when $m$ is odd (you didn't consider the parity), $p\mid x+y, p\nmid x,y$ (considering $p\neq 2$, which you haven't considered too, and you didn't consider $p\nmid x,y$, etc.). It is very unclear to me how you're using LTE here. – user26486 Feb 19 '15 at 15:25
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  • Is a direct consequence of Zsigmondy's theorem. If $x\neq y$, then, since $m\ge 2$, $x^m+y^m$ has a prime divisor that does not divide $x+y$. Since $x+y$ also has a prime divisor, we know that $x^m+y^m$ has at least two prime divisors, so $z$ cannot be prime.
    • – user26486 Feb 19 '15 at 15:35
  • Forgot to say $m$ is odd but I had intended to say this since $x+y$ only divides $x^m + y^m$ when $m$ is odd. Also $p$ doesn't divide $x$ or $y$. – Larry Freeman Feb 19 '15 at 16:03
  • Thanks for the reference to Zsigmondy's theorem. I'll read up on it. – Larry Freeman Feb 19 '15 at 16:05
  • The abc-conjecture is also related to equations such as this, although I see no straight forward application here. – Slime Online Feb 19 '15 at 16:12
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    I'm thinking you didn't quite state the problem correctly, since a counter-example is: n=1, x=2, y=2, m=3, z=16, p=2 – Gregory Grant Feb 19 '15 at 20:34
  • You are right. I think that other argument assumes that $p \ge 5$. I thought I could switch that to $x \ge 5$ which is not correct. I'll add that to the question. – Larry Freeman Feb 19 '15 at 21:10