If $p$ is an odd prime and $p < x < y < z$ are integers such that $$x^{p} + y^{p} = z^{p},\ \ p \mid y,$$ would it hold that $$z-x = p$$

If possible, I wish to know whether there is any paper proving this.

  • $\begingroup$ Note that if we assume we have a solution $x_0^p + y_0^p = z_0^p$ and put $x=Qx_0, y=Qy_0,z=Qz_0$ then $x^p + y^p = z^p$ is also a solution for any $Q$ including $Q=p$. So the answer is no - it does not imply $z-x=p$. $\endgroup$ – Winther Jun 21 '14 at 15:56
  • $\begingroup$ Thank you very much, Winther. But how is $z_{o} - x_{o}$? $\endgroup$ – Megadeth Jun 21 '14 at 15:59
  • $\begingroup$ You might also want to ask: if $p|y$, but $p\not| z,x$ do we have $p | (z-x)$. The answer here is yes as $0 \equiv y^p - z^p \equiv z-p$ mod $p$ from Fermats little theorem. So in general all you can say is that $z-x$ is divisible by $p$. $\endgroup$ – Winther Jun 21 '14 at 16:02
  • $\begingroup$ Ah Ha, thank you. But I myself have observed these facts. Indeed, Stewart in 1977 published in Mathematika the paper ``A note on the Fermat equation'', which gives the general form of $x+y, z-x, z-y$ under rather weak assumptions. The problem is I cannot get this paper for now and I cannot recall his results well enough at the present! $\endgroup$ – Megadeth Jun 21 '14 at 16:06
  • $\begingroup$ oai.cwi.nl/oai/asset/7429/7429A.pdf $\endgroup$ – Winther Jun 21 '14 at 16:40

I'll play around and see what your assumption of $p|y$ can lead to.

If $p | y$, then $p | z^p-x^p$. Letting $z = x+w$, $z^p-x^p =(x+w)^p-x^p =\sum_{k=0}^p \binom{p}{k} x^k w^{p-k} - x^p =\sum_{k=0}^{p-1} \binom{p}{k} x^k w^{p-k} $. Since $p | \binom{p}{k}$ for $1 \le k \le p-1$, $p | w^p$ so $p | w$.

Let $w = p^mv$ where $m \ge 1$ and $p \not \mid v$.

I will assume that $p \not \mid x$, or else $p \mid z$ and we could divide $p$ out until $p \not \mid x$ or $p \not \mid y$.

$\begin{array}\\ y^p &=\sum_{k=0}^{p-1} \binom{p}{k} x^k w^{p-k}\\ &=\sum_{k=0}^{p-1} \binom{p}{k} x^k (v p^m)^{p-k}\\ &=\sum_{k=1}^{p} \binom{p}{k} x^{p-k} (v p^m)^k\\ &=\sum_{k=1}^{p} \binom{p}{k} x^{p-k} v^k p^{mk}\\ &=p^m\sum_{k=1}^{p} \binom{p}{k} x^{p-k} v^k p^{m(k-1)}\\ &=p^m(x^{p-1}v+\sum_{k=2}^{p} \binom{p}{k} x^{p-k} v^k p^{m(k-1)})\\ \end{array} $

The first term ($x^{p-1}v$) is not divisible by $p$ and all the terms in the sum are, so $p^m \mid \mid y^p$ (i.e., $p^m \mid y^p$ and $p^{m+1} \not \mid y^p$ ). If $y = p^a b$ where $p \not \mid b$ (I'm running out of letters) $p^m \mid \mid y^p$ means $p^m \mid \mid p^{ap}b^p$ so $m = ap$ or $p \mid m$.

Therefore $p^p \mid y$.

However, I do not see how to take this further and, in particular, I do not see how this can imply that $p = w$.

  • $\begingroup$ You'd also need to assume that $x$, $y$ and $z$ are relatively prime, otherwise you could just multiply the whole by $p^n$ and have that divide the difference $\endgroup$ – vonbrand Jun 21 '14 at 17:46
  • $\begingroup$ Yep, I do. (characters added to make it happy) $\endgroup$ – marty cohen Jun 21 '14 at 21:13

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