Find whether there are integer solutions to $7x^2+9y^2=3932$. Is my solution correct? Find whether there are integer solutions to $7x^2+9y^2=3932$
Here's my attempt, I would like to know if it's correct please:
Lets assume $7x^2+9y^2=3932$ has integer solutions. That mean GCD of $x^2,y^2$ is 3932, or $(x^2,y^2)=3932$.
That means that there are $a,b \in \mathbb Z$ so that $x^2=3932a$, $y^2=3932b$.
And we know that $7x^2+9y^2=3932 \rightarrow 7\cdot 3932a+9\cdot 3932b=3932$
That means $7a+9b=1$, and since $(7,9)=1$, we can find such $a,b \in \mathbb Z$, using Euclid algorithem:
$9=1 \cdot 7 +2$
$7=3 \cdot 2 +1$
$2=2 \cdot 1+0$
Therefore, after calculations, I found:
$1=4 \cdot 7-3\cdot 9$
So $a=4$, $b=-3$. But we defined $y^2=3932b$, so we get $y^2$ to be negative. Therefore contraditon. 
Is this a valid contradiction? 
Thanks in advance for any assistance!

Notice that $7x^2 + 9y^2 = 67$ has a solution $(x,y) = (2,1)$. But this does not give that $\gcd(x^2,y^2) = 67$ here. So unfortunately, your reasoning does not hold.
On the other hand, if you reduce mod 3, you get $x^2 \equiv 2 \pmod 3$. The only squares mod 3 are $0$ and $1$, so there no solution.
