Show $15x^{2} - 7y^{2} = 9$ has no integer solutions

Solution 1:

Perhaps a little easier and shorter, work modulo $\,5\,$:

$$9=15x^2-7y^2=-7y^2=3y^2\Longrightarrow y^2=3$$

and it's easy to see that $\,3\,$ is a quadratic non-residue modulo $\,5\,$ , so we're done.

Solution 2:

You are not squaring the whole term, that is why you get stuck. Here is a solution. Clearly, since $7y^2=15x^2-9$ we get that $3|7y^2$ implying $3|y^2$ implying $3|y$. We get $y=3y_1$. We rewrite the equation as $15x^2-7(3y_1)^2=15x^2-63y^2=9$. Dividing by 3 both sides gives us $5x^2-21y^2=3$. This means $5x^2=3+21y^2$. Since RHS is divisible by 3, LHS is also divisible by 3. This tells us $3|x$. We get $x=3x_1$. Therefore, it follows that $45x_1^2=3+21y^2$. implying $15x_1^2-7y_1^2=1$. A consideration in modulo 3 reveals that, since $7y_1^2$ can never be $2 \;mod\;3$, the equation cannot have solutions in $\mathbb{Z}$.