The positive integer solutions for $2^a+3^b=5^c$ What are the positive integer solutions to  the equation

$$2^a + 3^b = 5^c$$ 

Of course $(1,\space 1, \space 1)$ is a solution.
 A: Another solution is
 $2^4 + 3^2 = 5^2$.  That's probably all.
A: 
case : $b>2$ and $c>2$

A calculus with computer, modulo $N=2372625=5^3\times 19 \times 37 \times 3^3$, in $H=\mathbb Z/N\mathbb Z  $ give 
$\text{card}(<2>)=900$, $\text{card}(3^3\times <3>)=900$, $\text{card}(5^3\times<5>)=36$.
And $(<2>+(3^3\times <3>)) \cap 5^3\times <5>=\emptyset$
with $1\in <g>$ the subset of $H$ generated by $g$ with times ($\times$)
So if $b>2$ and $c>2$, there are not solutions.

case : $b\leq 2$ and $c>2$

Here we choose $N=5^3\times (2^{25}-1)$
so $(<2>+\{1,3,9\}) \cap 5^3\times <5>=\emptyset$
So if $b\leq 2$ and $c>2$ no solution.

case : $c \leq 2$ three solutions

Then there are only 3 solutions, not any more.
A: If $a=0$, then it is clear there are no solutions. If $b=0$, then we need $2^a + 1 = 5^c$. It is easy to show in this case $a=2,c=1$ is the only solution by showing that we need $2^{a-2}|c$. When $c=0$ there are obviously no solutions.
Suppose $a=1$. Then $2 + 3^b = 5^c$ only has the solution of $b=1,c=1$. To show this check modulo $275$ to deduce $c=1$ and thus $b=1$ is forced.
Now, suppose $a \ge 3$. Then remark that by checking modulo $4$ we need $b$ to be even so let $b = 2b'$. So let's solve $2^a + 3^{2b'} = 5^c$. Checking modulo $8$ we get $c$ is even so let $c = 2c'$. Then: $$2^a = (5^{c'} - 3^{b'})(5^{c'}+3^{b'})$$
We get $5^{c'} - 3^{b'} = 2^m, 5^{c'}+3^{b'} = 2^n$ for some $m,n$.
But then $2 \cdot 5^{c'} = 2^m + 2^n$, forcing $m=1$. Thus we need $5^{c'} - 3^{b'} = 2$. But we already showed this only has the solution $b' = 1, c' = 1$. Thus it follows the only solution where $a \ge 2$ is with $a=4, b = 2, c= 2$.
Putting every together, we have proven the only solutions are $(2,0,1), (1,1,1), (4,2,2)$
EDIT: I realize I forgot to do the case of $a=2$. So we need to solve $4 + 3^{2b'} = 5^c$. Modulo $275$ happens to work again to force $c=1$ and thus we get no solutions when $b$ is nonzero.
A: Here is a little program in Python:
for k in range(10):
for l in range(10):
    for m in range(10):
        if 2**k + 3**l == 5**m:
            print (k, l, m)

Its output is as follows
    python check.py
1 1 1
2 0 1
4 2 2

This yields these three equations.
$$2 + 3 = 5$$
$$2^2 + 3^0 = 5^1$$
$$2^4 + 3^2 = 5^2.$$
And that's all folks!   Swap 100 for 10 and the result is the same.  Cooking up
an analytical proof of we have done is not hard.
A: 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.
