using binomial theorem variation $a, b, n, m\hspace{2mm}\epsilon\hspace{2mm}\mathbb{N}$ and 
$|n-m|$ is even (in other words $n,m$ are either both odd or both even)
Suppose for some $x \hspace{2mm} \epsilon\hspace{2mm}\mathbb{Z}\hspace{2mm}$, $x|a+1$ and $x|b+1$
Show that $x\hspace{2mm}|\hspace{2mm}a^n-b^m$
My solution:
Suppose $A = a+1$ and $B = b+1$
$$(A-1)^n - (B-1)^m$$
$$=\sum_{k=0}^{n} \binom{n}{k} A^{n-k} -\sum_{k=0}^{m} \binom{m}{k}B^{m-k}$$
both odd, then the last term of both sums will be $-1$. This will make the $-1,1$ cancel from both sums resulting in a sum of terms that are multiplied by A and B which are both divided by x and so the statement above is correct.
I'm just wondering if there's a better way to state my conclusion or conditions.
 A: Note that $a \equiv -1 (\bmod{x})$ and $b\equiv -1 (\bmod{x})$.  (Assuming $x$ is positive.  If not replace it by $-x$.)  Since $m$ and $n$ have the same parity, we have $a^n -b ^m \equiv (-1)^n - (-1)^m \equiv 0 (\bmod{x}).$
A: Put $\ c = -1,\,\ k = 2\ $ below, where we use standard Congruence Arithmetic Rules.
Lemma $\ $ If $\ k\mid n\!-\!m\ $ then $\ {\rm mod}\ x\!:\,\  a\equiv c\equiv b,\,\ \color{#c00}{c^k\equiv \bf 1}\,\Rightarrow\, a^n\equiv b^m$
Proof $\ $ Note $\,\ n = m\!+\!kj\ $  so  $\ a^{\large n}\!\equiv c^{\large n}\! \equiv c^{\large m}(\color{#c00}{c^{\large k}})^{\large j}\!\equiv c^{\large m}\color{#c00}{\bf 1}^{\large j}\!\equiv c^{\large m}\!\equiv b^m\pmod x $

Remark $ $ Essentially it boils down to  $\ c^{\large k}\equiv 1\,\Rightarrow\, c^{\large n}\equiv c^{\large m}\ \ $ if $\ n\equiv m\pmod k$
i.e. exponents on $\,c\,$ can be interpreted $\!\!\mod \!k\,$ when $\,c^{\large k}\equiv 1.\,$ The innate algebraic structure  will be brought to the fore when one studies cyclic groups (and modules).
