What is it, in Mihailescu's proof of Catalan conjecture, that uses explicitly the fact that there is a 1 on the right hand side of $x^p - y^q = 1$? In other words, why can't we extend his argument to prove stuff about, say, $x^p - y^q = 2$?



Can we extend the proof of Catalan's conjecture?
[This](http://www.ams.org/journals/bull/2004-41-01/S0273-0979-03-00993-5/S0273-0979-03-00993-5.pdf) article by Tauno Metsänkylä gives a good explanation of Mihăilescu's proof. There are some crucial steps in the proof that could not be done in the case $x^p-y^q=2,$ say. The idea is to write
$$\frac{x^p-1}{x-1}(x-1)=y^q$$
and observe that if $x-1$ and $\frac{x^p-1}{x-1}$ are coprime, we arrive at
$x-1=a^q, \quad \frac{x^p-1}{x-1}=b^q$
for some integers $a$ and $b$ (this could not be done with the equation $x^p-2=y^q$ as $x-2$ does not usually divide $x^p-2$). This is the so called first case that was solved by J. W. S. Cassels in 1960. Hence Mihăilescu was left with the second case where $\gcd(\frac{x^p-1}{x-1},x-1)=p$, so 
$$\frac{x^p-1}{x-1}=pb^q.$$
Denoting by $\zeta$ a primitive $p$th root of unity, we get
$$\prod\_{k=1}^{p-1}\frac{x-\zeta^k}{1-\zeta^k}=b^q$$
(again $\frac{x^p-2}{(x-2)p}$ does not have such a nice factorization).
By moving to ideals, we get the equation
$$\left\langle\frac{x-\zeta^k}{1-\zeta^k}\right\rangle=J^q,$$
where $J$ is an ideal of $\mathbb{Z}[\zeta]$. One of Mihăilescu's key ideas was to transfer this equation back to an equation of numbers by using an annihilator $\theta$ of $\mathbb{Z}[\zeta]$ to arrive at
$$\left(\frac{x-\zeta^k}{1-\zeta^k}\right)^{\theta}=\varepsilon \gamma^q,$$
where $\varepsilon$ is a unit in $\mathbb{Z}[\zeta]$ and $\gamma\in \mathbb{Q}(\zeta)$. From this equation, he was able to derive his miraculous congruences
$$x\equiv 0 \pmod{q^2},\quad p^{q-1}\equiv 1 \pmod{q^2}, \quad y\equiv 0 \pmod{p^2},\quad q^{p-1}\equiv 1 \pmod{p^2},$$
which are a key to the proof. The equation $x^p-y^q=2$ may actually be an open problem; at least I could not find any references on it. But of course if we accept Mochizuki's proof of the abc-conjecture, it has only finitely many solutions.


