Estonia — Open Contests

Let $d = \gcd(a+b, c)$. Since $c \mid a+2b$, also $d \mid a+2b$. Now $(a+2b) - (a+b) = b$ and $2(a+b) - (a+2b) = a$ are divisible by $d$. Therefore $d$ is the common divisor of $a$, $b$, $c$ and due to our initial assumption of $a$, $b$, $c$ being relatively prime it has to be $1$. Hence $a+b$ and $c$ are also relatively prime. But since $a^2 - b^2 = (a-b)(a+b)$ is divisible by $c$, the factor $a-b$ has to be divisible by $c$.

Since $a+2b$ is divisible by $c$, also $(a-2b)(a+2b)$ is divisible by $c$. But $(a-2b)(a+2b) = a^2 - 4b^2$; since $a^2 - b^2$ is divisible by $c$, the differences $(a^2 - b^2) - (a^2 - 4b^2) = 3b^2$ and $4(a^2 - b^2) - (a^2 - 4b^2) = 3a^2$ also have to be divisible by $c$.
If $3 \nmid c$, then $c \mid a^2$ and $c \mid b^2$. Therefore every prime divisor of $c$ would also be a prime divisor of $a$ and $b$, which would contradict the initial assumption. Hence either $c = 1$, in which case the problem statement holds trivially, or $3 \mid c$. In the latter case let $c = 3c'$; the statements above show that $c' \mid a^2$ and $c' \mid b^2$. As we saw above, every prime divisor of $c'$ would be a prime divisor of $a$ and $b$, due to which $c' = 1$ and $c = 3$. Now since $a+2b$ and $3b$ are divisible by $3$, also $(a+2b) - 3b = a - b$ is divisible by $3$.