Solve the diophantine equation: $5^m + n^2=3^p$ where $m,n,p \in \mathbb{N}-\{0\}$
One solution is $m=1,n=2,p=2$. Now, applying modulo 4:
$1 + 0 = (-1)^p \mod 4 \tag 1$ or $1 + 1 = (-1)^p \mod 4 \tag 2$
But only (1) is possible, therefore both $n, p$ are even.
Also, applying modulo 3:
$(-1)^m + n^2 = 0 \mod 3 \tag 3$ therefore $m$ is odd and $n=3k \pm 1$
I could not get any further.
Following lulu, to prove there are no solution with $p\geq 3$, we only need to prove the equation $$2(3^k) - 5^m = 1$$ has no solution with $k\geq 2$.
Note that $5$ is a primitive root modulo $3^2$, hence $5$ is a primitive root modulo $3^n$ for all positive integer $n$. Thus $$5^m \equiv -1 \pmod{3^{k}} \implies \frac{\phi(3^{k})}{2}\mid m \implies 3^{k-1}\mid m $$ Therefore $(1)$ implies, for some $r\geq 1$ $$1 = 2({3^{k}}) - 5^{r(3^{k-1})} \leq 2({3^{k}}) - {5^{{3^{k-1}}}}$$ a contradiction, because RHS is negative when $k\geq 2$.
Therefore, the only positive integer solution of $5^m + n^2=3^p$ is $m=1,n=2,p=2$.