Find all positive integer solutions for the following equation $5^a + 4^b = 3^c$: 
Find all positive integer solutions for the following equation:
$$5^a + 4^b = 3^c$$

My first guess would be to study the equation in mod, but I tried modulo 3, 4, 5, and 9 and I can't find anything.
 A: Hint: $$1^a + 0 \equiv _4 (-1)^c \implies c \equiv_2 0$$
so $c=2d$, $d\in \mathbb{N}$.
So $$ 5^a = (3^d-2^b)(3^d+2^b)$$
so $3^d-2^b = 5^x$ and $3^d+2^b = 5^y$. Now add/substract both equations...


...we see that $x=0$. Now if $b>1$ then we have $$(-1)^d -0 \equiv _41 $$ so $d$ is even. So we have $$(3^m-1)(3^m+1)= 2^b$$ Now $(3^m-1)$ and $(3^m+1)$ are two consecutive even numbers so exactly one is divisible by $4$ so the other number is exactly $2$.

A: Another route: Follow the reasoning in Aqua's answer up to $5^a=(3^d-2^b)(3^d+2^b)$.
Then ask if $3^d-2^b$ and $3^d+2^b$ share any factors. If $m\mid (3^d-2^b) \land m \mid (3^d+2^b)$, then $m \mid (3^d-2^b)+(3^d+2^b)=2\cdot3^d$, which means $m\in\{1,2,3^x\}$. But $2\not \mid (3^d-2^b)$ and $3\not \mid (3^d-2^b)$, so $m=1$. In other words, $\gcd((3^d-2^b),(3^d+2^b))=1$
But the product of the two terms is a power of $5$, so one term must be exactly $1$ and the other term is the power of $5$. Plainly the smaller term will be $1$, so we solve $(3^d-2^b)=1$, which has only two solutions: $b=d=1$, and by Mihailescu's theorem $d=2,b=3$
Substituting into the other term, we find $3^1+2^1=5^1$ and $3^2+2^3=17 \ne 5^d$. So there is only one solution to the original equation: $5^1+4^1=3^2$
