Prove that there are only 3 integer solutions to $2^a+3 ^b=5^c$ I have figured out that $a \pmod{2} = b\pmod{2} =c \pmod{2} $ by making a few modulo tables and using that both sides of the equation must be divisible by $5$, and I have found the three solutions, I'm just at a loss as to how to prove that those are the only three. I struggle specifically with how to move from something being divisible by $5$ to something being a power of $5$. I have also tried to use the binomial expansion by noting that $5=2+3$, but I quickly ran out of ideas what to do with it. 
Any help would be appreciated! 

There are three solutions which can all be found by elementary means.
If $b$ is odd 
$$2^a+3\equiv 1 \bmod 4$$
Therefore $a=1$ and $b$ is odd.  
If $b>1$, then $2\equiv 5^c \bmod 9$ and $c\equiv 5 \bmod 6$ 
Therefore $2+3^b\equiv 5^c\equiv3 \bmod 7$ and $b\equiv 0 \bmod 6$, a contradiction. 
The only solution is $(a,b,c)=(1,1,1)$. 
If $b$ is even, $c$ is odd 
$$2^a+1\equiv 5 \bmod 8$$
Therefore $a=2$. 
If $b\ge 2$, then $4\equiv 5^c \bmod 9$ and $c\equiv 4 \bmod 6$, a contradiction. 
The only solution is $(a,b,c)=(2,0,1)$. 
If $b$ and $c$ are even 
Let $b=2B$ and $c=2C$. Then
$$2^a=5^{2C}-3^{2B}=(5^C-3^B)(5^C+3^B)$$
Therefore $5^C-3^B$ is also a (smaller) power of 2. 
A check of $(B,C)=(0,1)$ and $(1,1)$ yields the third solution $(a,b,c)=(4,2,2)$.
$(B,C)=(2,2)$ does not yield a further solution and we are finished.
