Problem 12.1. Solve in integers the equation

$$
2^{a}+8 b^{2}-3^{c}=283
$$

Oleg Mushkarov, Nikolai Nikolov
12.1. It is easy to see that $a, c \geq 0$. Since $3^{c}$ is congruent to 1 or 3 modulo 8 , then $0 \leq a \leq 2$. If $a=0$ or $a=1$, then 2 divides $3^{c}$ or 8 divides $3^{c}+1$, a contradiction.

Let $a=2$, i.e., $8 b^{2}-3^{c}=279$. The cases $c=0,1$ are impossible and hence $c \geq 2$. Then 3 divides $b$ and setting $b=3 d$ gives $8 d^{2}-3^{c-2}=31$. If $c \geq 3$, then 3 divides $d^{2}+1$, a contradiction. Therefore $c=2$ and $d= \pm 2$, i.e. $a=2$, $b= \pm 6$ and $c=2$.