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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

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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.

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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 $\mathbb{Z}$.

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    $\begingroup$ Thanks, modulo 3 is the key I guess. Although, I'm not sure what you meant by "not squaring the whole term" as I got the same equation (15x^2 - 7y^2 = 1) as you. $\endgroup$ – scibuff Jan 11 '13 at 12:58
  • $\begingroup$ @scibuff my bad, I thought you were writing $(3y_1)^2=3y_1^2$ $\endgroup$ – mathemagician Jan 11 '13 at 13:01
  • $\begingroup$ @mathemagician how do you say that $7y_1^2$ can never be $2(mod \:3)$ $\endgroup$ – Umesh shankar Jun 6 at 17:07

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