Problem from Competition for JBMO

$7^{a} - 4^{d} - 6^{b} = 5^{c}$

If $b > 0$ , then the LHS is divisible by 3, which is a contradiction. That means $b = 0$ . We also notice that if $b = 0$ , $d$ also has to be zero, otherwise the LHS would be divisible by two. This gives us

$7^a = 5^c + 2$

Since the equation $7^a \equiv 2 \pmod{25}$ has no integer solutions $a$ , that means $c < 2$ . This gives us the only solution $(a, b, c, d) = (1, 0, 1, 0)$ .