Number or triples $(a,b,c)$ such that $2^a - 5^b*7^c =1$

Question

$$\large 2^a - 5^b7^c =1 $$

How many possible triples $(a,b,c)$ are possible?

I tried solving this modulo $5$ but that didn't help. Suggestions?

Answer 1

Hint

If $5|2^a-1$ then $4|a$ and hence $2^4-1|2^a-1$. Thus

$$3| 2^a-1=5^b7^c \,.$$

To complete the proof, you need to see solve the case $a=0$, which yields

$$2^a=7^c+1 \,.$$

$a=3$ is clearly a solution, and you show that there is no other one. [Mihailescu Theorem would prove this, but it is overkill].

Answer 2

This might be only an insight. Can we count (assuming finiteness) solutions to diophantine equation $2^m=35n+1$ (or equivalently $35k=1 \mod2$)? then the answer should be highly relevant to the triple $(a,b,c)$ of the given equation.