Find all the nonnegative integers $a,b,c$ such that $1+2^a = 4 \cdot3^b + 5^c$.
I found this problem in an old number theory problem set. Using a computer, I found that the only solutions for $a,b,c \leq 30$ are $(2,0,0),(3,0,1),(4,1,1),(7,0,3),(12,5,5)$. I'm tempted to say that these are the only ones, but I haven't been able to prove it.
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Observe that all the solutions appear to have $b \leq 5$. Treat the cases with $0 \leq b \leq 5$ as Will suggests (by local means or using lower bounds for linear forms in two logarithms), to reach the conclusion that the only solutions with $b < 6$ are the ones you have already found. Suppose that $b \geq 6$; note that both $2$ and $5$ are primitive roots mod $3^6$, having order $486$. We find precisely $243$ pairs of residue classes modulo $486$ for $(a,c)$ such that $$ 1+2^a \equiv 5^c \mod{3^6}. $$ Considering the original equation modulo various primes $p \equiv 1 \mod{486}$, we find that only the pair $a \equiv 248 \mod{486}$ and $b \equiv 427 \mod{486}$ is compatible with $1+2^a=4 \cdot 3^b+5^c$ modulo both $p=487$ and $p=1459$. Considering the equation modulo $p=17497$ then provides a contradiction.