Let $n,p,q$ be positive integers. Solve the diophantine equation $$(n+2)^p-n^q=2.$$
I known this is Catalan's Conjecture, and the special case $n=3$ ($5^m = 2 + 3^n$ help what to do) has simple methods.
But I ask this question: have simple methods(such $n=3$)?
because it is contest math problem
If $4 \mid n$, we can quickly eliminate all $p\ge2$; when $2 \mid n$, the options are very finite and easily calculated/determined.
Now assume $n$ is odd. Expanding the first term yields $2^p$ plus a whole bunch of terms with a factor of $n$. Therefore, we can reduce the equation modulo $n$ to find $$2^p \equiv 2\!\!\!\!\pmod{n},$$ and since $n$ is odd, this implies $$2^{p-1} \equiv 1\!\!\!\!\pmod{n}.$$ This equation is well represented in the literature, and there are elementary and relatively simple methods available to attack/solve it.