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
 A: 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.
A: Hint: If $b > 49$, you know $b' \geq 25$. But $b' \geq 30$ is already invalid because $2\cdot 30^2 \geq 1694$. Try the rest.
A: Notice that
$$ 3a+b^{2} = 6a' + 4(b')^{2} = 3388 \implies 3a' + 2(b')^{2} = 1694 $$
So we have
$$ (b')^{2} =\frac{1694-3a'}{2}$$
From here, $a'$ must be even, $a'=2m$. Since $a'$ is even, then $b'$ must be odd, because if not..then $4|a$ and $4|b$ which is contradiction. Now we have
$$ (b')^{2} = 847-3m$$
and the choices are $b'= 25, 27$, or $29$..
A: We can have  the general solution for the equation $3a+b^2= 3388$ in positive integers, given that $\gcd(a,b)=2$.
Indeed, writing as above $a=2a'$, $b=2b'$, where $\gcd(a',b')=1$, we obtain 
$$3a'+2b'^2=1694,$$
which shows first that $a'$ is even, and second, that $\;2b'^2\equiv 2\mod 3$, hence $b'\equiv 0\mod 3$. 
So write $a=2a''$ to obtain the equation 
$$3a''+b'^2=847,\quad b'\not\equiv 0\mod 3.$$
Furthermore, $b'^2 <847$, hence $b'\le 29$. 
Conversely, if an odd $b'\le 29$ is not a multiple of $3$, then $b'^2\equiv 1\mod 3$, so $847-b'^2\equiv 1-1=$$0\mod 3$, and we can compute the value of $a''$, whence $a=4a''$ and $b=2b'$.
There are $10$  values for $b'$ which satisfy these conditions:
$$\{1,5,7,11,13, 17, 19,23,25,29\}$$
As an example, take $b'=19$, so $3a''+361=847$, whence $a''=\frac13 486=162$, and finally
$$a=648,\enspace b =38, \quad 3a+b^2=1944+ 1444=3388.$$
For the supplementary condition in the O.P.'s question ($b\ge 49$) you have to retain th values $b'=25$ and $b'=29$.
