Proving that an equation doesn't have integer solutions

Question

I need to prove that there are no integer solutions for a bunch of equations like the following: $$15x^2 - 7y^2 = 9$$ I was able to solve some simpler ones by picking a dividend and looking into it's remainder table. But it's not working for the others. How should I start thinking this kind of problem? It's from my algebra class and we are looking into divisibility and congruence. Thanks!

Answer 1

Hint: Maybe note first that $3$ must divide $y$. Let $y=3t$. Then we are looking at $15x^2-(7)(9t^2)=9$. So $3$ must divide $x$. Let $x=3s$. We end up with $$15s^2-7t^2=1.$$ Now work modulo $3$.

Answer 2

If $15x^2-7y^2=9$ for some $x,y\in\mathbb Z$, then $-7y^2\equiv 9\pmod{5}$, i.e. $$3y^2\equiv 9\pmod{5}\stackrel{:3}\iff y^2\equiv 3\pmod{5},$$

contradiction, because $3$ is not a quadratic residue mod $5$. To prove this, notice $(5k)^2\equiv 0\pmod{5}$, $(5k\pm 1)^2\equiv 1\pmod{5}$, $(5k\pm 2)^2\equiv 4\pmod{5}$ for all $k\in\mathbb Z$.

Answer 3

Call $x^2=a$ and $y^2=b$ then we have $15a-7b=9$ which is a diaphante equation. It's solution can be found by Euclid's algorithm, but also there is a theorem that no solutions exist iff $9$ isn't a multiple of the $gcd$ of $15$ and $7$.

In this case there are solutions, so you can find them, then show if they can be or can't be 2 squares.

Answer 4

You have modulo $5$ $$-2y^2\equiv 4\pmod 5\iff y^2\equiv -2\pmod 5\iff y^2\equiv 3\pmod 5$$ But the only squares modulo $5$ are $0,1$ and $4$ hence there are not solutions.