prove that there are no natural numbers for which $15x^2-7y^2=9$ [duplicate]

my approach:

$15x^2-7y^2=9\implies y^2=(15x^2-9)/7$

$$\implies (14x^2-7+x^2-2)/7=(2x^2-1)+(x^2-2)/7$$

then I found that $x^2-2$ is divisible by $7$ only when it is of the form $7n\pm3,$

so then replaced $x$ by $7n\pm3$ over here $y^2=(15x^2-9)/7.$

so $y^2=15(49n^2\pm42n+7).$

here $y^2$ is possible only when both the roots are equal which is possible only when the discriminant is $0$ ,so i tried and got the discriminant as $540$. so there is a non-zero discriminant,so no solution is possible. Is my proof correct or am i going wrong ?

Solution 1:

continuing from $y^2=15(49n^2 \pm 42n+7)$,
$3\mid 49n^2+7$
$3\mid 7n^2+1$
$n^2\equiv 2\pmod{3}$
contradiction

Solution 2:

$15x^2-7y^2=9$.

3 must divide y since it divides all but the y term and doesn't divide 7. This implies the term with $x$ is divisible by 9, but 15 is only divisible by 3. It follows that x is also divisible by 3. Let y=3q and x=3p.

$135p^2-63q^2=9$

$15p^2-7q^2=1$

$3q^2\equiv 1 \pmod{5}$

2 is the multiplicative inverse of 3 modulo 5.

$q^2 \equiv 2 \pmod{5}$

There is no such $q$, so we have a contradiction.