This is the equation for which I've to show that there are no integer solutions:
$5x^2-2y^2=4$
Any approach works! I’m just confused because I see some people approach by showing hyperbolas and some approach with a more modular approach.
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1So you need to choose your $n$ carefully. Since one of the coefficients is $5$, $n=5$ can be tried : does $y^2=-2$ has a solution modulo $5$? – Sarvesh Ravichandran Iyer Oct 19 '20 at 03:19
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Also the hyperbolic approach, well you must tell us where you saw it, and we can tell you when it applies. – Sarvesh Ravichandran Iyer Oct 19 '20 at 03:34
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Well, it's the converse. If there's no solution in mod $n$, then there's no solutions in integers. Because $x^2=2$ doesn't have integer solutions, but $x^2\equiv2\pmod 7$ has solution $x\equiv3,4$.
Since $5$ is one of the coefficients, it might be worth to try modulo $5$, because the $x^2$ is destroyed. Then $-y^2\equiv2\pmod5$, or $y^2\equiv3\pmod5$. However, all quadratic residues (values $n$ such that there's an $a$ for $a^2\equiv n$) in mod $5$ are $0,1,4$. Then there's no solution in integers.
dua
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