$3^n - 2^m = \pm 41$ is not possible. How to prove it? $3^n - 2^m = \pm 41$ is not possible for integers $n$ and $m$. How to prove it?
 A: The congruence $3^n - 2^m \equiv 41\pmod{60}$ has no solutions.
The congruence $3^n - 2^m \equiv -41\pmod{72}$ has no solutions.
A: As a valuable hint for solving the problem, I consider the following extract from my lectures on elementary number theory.
Theorem ($\approx1320$; Levi ben Gerson 1288--1344).
The equations
$$
(1) \quad 3^p-2^q=1
$$
and
$$
(2) \quad 2^p-3^q=1
$$
have no solutions in integers $p,q>1$, except the solution
$p=2$, $q=3$ to equation (1).
Proof.
(1) If $p=2k+1$, then
$$
2^q=3^p-1=3\cdot9^k-1\equiv2\pmod4,
$$
which is impossible for $q>1$.
If $p=2k$, then $2^q=3^p-1=(3^k-1)(3^k+1)$ implying
$3^k-1=2^u$ and $3^k+1=2^v$. Since
$2^v-2^u=(3^k+1)-(3^k-1)=2$, we have $v=2$ and $u=1$.
This corresponds to the unique solution
$q=u+v=3$ and $p=2$.
(2) If $q\ge1$, then $3^q+1$ is not divisible by~$8$.
Indeed, if $q=2k$, then $3^q+1=9^k+1\equiv2\pmod8$;
and if $q=2k+1$, then $3^q+1=3\cdot9^k+1\equiv4\pmod8$.
Therefore $p\le2$, hence $p=2$. The latter implies $q=1$
which does not correspond to a solution.
