Find the number of positive integer solutions to $2^a - 5^b \cdot 7^c = 1$.
By modular arithmetic arguments: $b$ must be even and $c$ must be odd; $a$ must be divisible by 4. But then modulo 3: $(-1)^a - (-1)^b \equiv 1 \pmod{3}$ gives $1 - 1 \equiv 1 \pmod{3}$, i.e. $0 \equiv 1 \pmod{3}$, which is impossible. The equation has no solution. (D)
Find the number of positive integer solutions to $2^a - 5^b \cdot 7^c = 1$.