Is there a simple proof that there are no "off-by-1" solutions to Fermat's Last Theorem?

Question

$x^p + y^p = z^p$ 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.)

Answer 1

This is only a comment until I work out the even-$p$ case. Work modulo $x+1$ so $(-1)^p \equiv 1$. This refutes all cases with odd $p$. If $p$ is even $a=x^q,\,b:=(x+1)^q,\,c:=(x+2)^q$ is a Pythagorean triple with $q:=p/2$. I'm still trying to force a contradiction. Since $c-b>b-a$, $a^2=c^2-b^2>(c+b)(b-a)>(3b-a)(b-a)=a^2+4b(b-a)-b^2>a^2-b^2$.

Answer 2

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.

Answer 3

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.