Solve $\gcd(a,b)=2$, $3a+b^2 =3388$, $a>0$, $b>49$

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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 $29$..

Answer 4

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 $b'=29$.