# 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.

Thanks in advance for any assistance!

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Why does the first assumption imply that $x^2$ and $y^2$ have gcd $3932$? – user61527 Mar 7 '14 at 22:33
Personally I would look at this modulo $3$ – Mark Bennet Mar 7 '14 at 22:34
@T.Bongers to amplify $7\cdot 2+ 9 \cdot 3=41$. That doesn't mean the GCD of $2$ and $3$ is $41$. – Mark Bennet Mar 7 '14 at 22:36
I see... thank you. When does it mean that though? I wasn't taught this with much rigor... – charlie Mar 7 '14 at 22:37
One cannot (at least as I see) immediately conclude that $x^2$ and $y^2$ have no common factor: What we can conclude, however, is that the $(x^2, y^2)$ is a divisor of $3932$. More generally, any two linear combination of two numbers (in this case, a linear combination of $x^2$ and $y^2$) must be divisible by the GCD of the two numbers, but no more is true. @MarkBennet's suggestion of working modulo $3$ will help a lot, since we can immediately ignore the variable $y$ and only look at $x$. – user61527 Mar 7 '14 at 22:39

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.
Why look mod $3$? - always useful to try small primes to see what might be going on. $3$ eliminates on term too, so simplifies things. If that didn't work I'd try mod $7$. And if that didn't work I'd have some clues as to how a solution might look. $2$ would be another prime to choose - but you might try mod $4$ or mod $8$. This doesn't always work, and there are other things to try - but it is cheap and easy if you can't see where else to begin. – Mark Bennet Mar 7 '14 at 23:19