Solve $\gcd(a,b)=2$, $3a+b^2 =3388$, $a>0$, $b>49$ I have this problem and I can't do it.
$$\begin{cases}
\gcd(a,b)=2 \\ 
3a+b^2 =3388 \\ 
a>0,b>49
\end{cases}
$$
I've tried writing $a=2a'$ and $b=2b'$, but then I have $3a'+2b'^2=1694$ and I don´t know what to do

Assuming $a$ needs to be positive, the equation $3a + b^2 = 3388$ shows that $b \leq 58$.  Since $gcd(b,a) =2 $ implies that $b$ is divisible by $2$, so the only options are $b = 50,52,54,56,58$.  Note that $4$ divides $3388$, so if $4$ divides $b$, then $4$ divides $3a$ and hence $4$ divides $a$.  This is not acceptable, since $gcd(b,a)$ is required to be $2$.  Thus we find $b \neq 52,56$.  
Examining $3a + b^2 = 3388$ modulo $3$, we see that $b^2 \equiv 1 \mod 3$.  Thus $b \equiv 1, 2 \mod 3$ are the only possibilities.  Hence $b$ is not $54$.  
Now the only remaining possibilities are $b = 50, 58$, which you can check are both solutions.
