<s>[INST] Is there a simple proof that there are no "off-by-1" solutions to Fermat's Last Theorem? $x^p + y^p = z^p$ has no integral solutions for $p>2$ due to the work of Andrew Wiles and others. My questions is: consider cases where $p>2$ and $y=x+1$ and $z=x+2$. We might call these "off-by-1" solutions. Is there any simple proof that there are no such off-by-1 solutions? (Admittedly, simple is somewhat subjective.)
 [/INST] Let's anchor it at the middle term, so that we're looking for integers $p > 2$ and $y$ such that
$$(y - 1)^p + y^p = (y + 1)^p. \tag{1}$$
J.G. elegantly disposed of the case where $p$ is odd, then reducing $(1)$ modulo $y$ yields
$$-1 = (-1)^p \equiv (y - 1)^p + y^p = (y + 1)^p \equiv 1^p = 1 \pmod{y}\,.$$
This congruence is satisfied only for $y \mid 2$, i.e. $y \in \{ -2,-1,1,2\}$, and one easily checks that none of these options gives a solution of $(1)$.
For even $p$, $y = 0$ yields the only solution. For $y < 0$ and even $p$ we have
$$(y - 1)^p + y^p = (\lvert y\rvert + 1)^p + \lvert y\rvert^p > \lvert y\rvert^p > (\lvert y\rvert - 1)^p = (y + 1)^p,$$
so it remains to rule out positive $y$. For $y \in \{1,2\}$ and $p > 1$ we always have
$$(y - 1)^p + y^p < (y + 1)^p,$$
hence in the following we assume $y > 2$. If $(1)$ were satisfied, then by the binomial theorem
$$y^p = (y + 1)^p - (y - 1)^p = 2\sum_{k = 1}^{p/2} \binom{p}{2k-1} y^{2k-1} = 2py + y^3\sum_{k = 2}^{p/2} \binom{p}{2k-1}y^{2k-4}\,.$$
The left hand side is divisible by $y^3$ (since $p > 2$), and the right hand side is divisible by $y^3$ if and only if $y^2 \mid 2p$. We thus obtain the necessary condition $p \geqslant \frac{1}{2}y^2$. But then we have
$$\biggl(1 - \frac{1}{y}\biggr)^p + 1 < 2 < 1 + \frac{y}{2} \leqslant 1 + \frac{p}{y} \leqslant \biggl(1 + \frac{1}{y}\biggr)^p$$
by Bernoulli's inequality.
Thus for $p > 2$ the equation $x^p + y^p = z^p$ has no nontrivial solutions with $x,y,z$ being consecutive integers.
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