<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] It is easy (and nice!) to discard $p$ odd (look at the J.G.'s answer) because of $(-1)^p\equiv 1\mod{(x+1)}$ impossible for $x\gt1$ and because the equation $1+2^p=3^p$ can be easily discarded so we consider just $p=4n$ and $p=4n+2$.
It is well known the periodicity modulo $4$ of classes modulo $10$. We have modulo $10$ the following table
$$\begin{array}{|c|c|}\hline x & 1 & 2 & 3 & 4&5&6&7&8&9&0 \\\hline x^2  & 1 & 4 &9&6&5&6&9&4&1&0\\\hline x^4 &1 &6  &1&6&5&6&1&6&1&0\\\hline\end{array}$$ It follows
► for $p\equiv 2\pmod4$.
$$\begin{cases}1^2+2^2=3^2\Rightarrow5\equiv9\pmod{10}\\2^2+3^2=4^2\Rightarrow3\equiv6\pmod{10}\\3^2+4^2=5^2\Rightarrow5\equiv5\pmod{10}\\4^2+5^2=6^2\Rightarrow1\equiv6\pmod{10}\\5^2+6^2=7^2\Rightarrow1\equiv9\pmod{10}\\6^2+7^2=8^2\Rightarrow5\equiv4\pmod{10}\\7^2+8^2=9^2\Rightarrow3\equiv1\pmod{10}\\8^2+9^2=0^2\Rightarrow5\equiv0\pmod{10}\\0^2+1^2=2^2\Rightarrow1\equiv4\pmod{10}\end{cases}$$
(This can be written directly from the table but has been made explicit for beginners).
We see that WLG we can proved just the impossibility of
$$(10m+3)^{4n+2}+(10m+4)^{4n+2}=(10m+5)^{4n+2}.....\space\space (*)$$
►for $p\equiv 0\pmod4$.
Similarly we should try only the equation
$$(10m+5)^{4n}+(10m+6)^{4n}=(10m+7)^{4n}.....\space\space (**)$$
So we have reduced the problem to just proving  $(*)$ and $(**)$. I stop here until a new attempt and maybe someone can prove beforehand the impossibility of these two equations for which the Pythagorean triples could be perhaps a good way.
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