TITLE: Show $15x^{2} - 7y^{2} = 9$ has no integer solutions
QUESTION [12 upvotes]: I'm trying to show the quadratic binary has no integer solution. I've used the following process to transform it into a Pell's equation of the form $x^{2} - Dy^{2} = M$
If there is a solution, then $3\mid y$ so we can write $y = 3y_{1}$ and the equation becomes
$3(5x^{2})-7(3y_{1}^{2}) = 3(3)$ which is just
$5x^{2}-21y_{1}^{2} = 3$
now I believe I can just the same process again taking $x = 3x_{1}$ to get 
$5(3x_{1}^{2})-3(7y_{1}^{2}) = 1$ which is just
$15x_{1}^{2}-7y_{1}^{2} = 1$
Finally, multiplying by 15, taking $X = 15x_{1}$ and $Y = y$ we get
$X^{2} - 105Y^{2} = 15$
and now I'm stuck as I can't find a reason why this shouldn't have a solution.
Thank you

REPLY [26 votes]: 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.

REPLY [13 votes]: 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 $\mathbb{Z}$.