TITLE: Finding all solutions for $3^c=2^a+2^b+1$
QUESTION [9 upvotes]: 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?

REPLY [4 votes]: 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 $(4,6)$).