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I'm trying to show that:

$x^2 + 14y^2 = \pm 2$ has no integer solutions.

My initial thought was the usual approach of considering modulo 7, but this seems to fail. My next guess was to consider modulo 5, which yields $\pm 2 = (x+y)(x-y)$ but this seems to fail too. I'm not sure where to go next.

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  • $\begingroup$ Do you mean $x^2+14y^2$ or $x^2-14y^2$? The first is easy, since we must have $|x| \le 1$ and $y=0$. $\endgroup$ Mar 18, 2017 at 22:00

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Consider that for $x=0$ or $y=0$ there is no solutions and we have for $x,y\geq 1$ that $x^2+14y^2\geq 1+14\geq 15> 2$

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  • $\begingroup$ A nit: the last $\ge$ should be $\gt$. Otherwise one could posit that all the $\ge$s were $=$s. $\endgroup$ Mar 18, 2017 at 16:21

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