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

Question

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

Answer 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.

Answer 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 $\mathbb{Z}$.

Answer 3

$$15x^2 - 7y^2 = 9 \quad \implies \quad 15x^2 - 9 = 7y^2.$$

Now, it is obvious that LHS is divisible by 3, which means that RHS has to be divisible by 3 as well. This implies that $y^2$ is divisible by 3; thus is $y$ divisible by 3.

Let $E$ be the set of possible ending digits (unit digits) of RHS. Simple calculation gives

$$E = \{0, 2, 3, 5, 7, 8\}.$$

However, if we look at the equation $15x^2 - 9 = 7y^2$ we see that $7y^2 + 9$ has to be divisible by 15 in order for $x$ to have an integer solution. This means that $7y^2 + 9$ must end with a 5 or a 0 in order for it to be divisible by 5 (since it has to be divisible by 15). This gives that the ending digit of $7y^2$ has to be 6 or 1, but we see that $6 ∉ E ∧ 1 ∉ E$, which means that $7y^2 + 9$ can't be divisible by 15. Hence the equation does not have any integer solutions.

$Q.E.D.$