Finding all solutions for $3^c=2^a+2^b+1$ Given the equation:
$3^c=2^a+2^b+1$
Find all solutions for $a,b,c$ - given that they are positive integers and $b>a$.
Any ideas? 
 A: This equation can be solved using Baker's method. It is explained in the book Unsolved problems in number theory by Richard Guy. He writes "In fact, neither $3^a=1 +2^b+2^c$ nor $2^a+3^b=2^c+3^d$ can be reduced to a single congruence. But Baker's method makes it possible to solve these last two (exponential) Diophantine equations." Then he gives the relevant bibliography where this has been done.
In fact, the solution is attributed to Pillai. All integer solutions of $3^a=1 +2^b+2^c$ are
$$
(a,b,c)=(1,0,0),(2,2,2),(4,6,4),(4,4,6).
$$
S.S. Pillai: On the equation $2^x-3^y=2^z+3^w$, Bull. Calcutta Math. Soc. $37$, 1945.
A: It's still relatively simple to handle such exponential equations using local arguments. If $a \geq 5$, then we have $3^c \equiv 1 \pmod{32}$ and hence $8 \mid c$. It follows that $3^c \equiv 1 \pmod{41}$ and so $2^{b-a} \equiv -1 \pmod{41}$. We therefore have $b-a \equiv 10 \pmod{20}$ which implies that $2^{b-a} \equiv -1 \pmod{25}$ and $3^c \equiv 1 \pmod{25}$, whereby $5 \mid c$ and $2^{b-a} \equiv -1 \pmod{11}$. This last congruence implies that $b-a \equiv 5 \pmod{10}$, contradicting the parity of $b-a$.
The cases where $a=0, 2$ and $4$ are (more) easily treated (leading to the solutions with $(a,b)=(0,0), (2,2)$ and $(4,6)$).
A: working in $\pmod3$ we find that 
$0=(-1)^a+(-1)^b+1 \pmod3$
$edit;$
it only shows that $a$ and $b$ must be even, in $\pmod4$ we see that $c$ also must be even
