Consider the following Diophantine equation $$ 2x^2-Ny^2 = -1. $$ where $N$ is an integer. Is there any result expressing the values of $N$ for which the above equation admits an integral solution?
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$\begingroup$ $-2$ has to be a square modulo $N$, so every prime dividing $N$ has to be $1$ or $3$ modulo $8$. $\endgroup$– Gerry MyersonCommented Oct 20, 2023 at 5:43
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1 Answer
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If you multiply this by $2$ you get $(2x)^2 - 2N y^2 = -2$, so you can solve $x^2 - 2N y^2 = -2$ and then check when $x$ is even. $x^2 - 2Ny^2 = -2$ is a generalized Pell's equation, so you can find a general form for its solutions, and then you can look at the general solution (in matrix form) mod $2$ and see if $x$ can ever be even. It doesn't seem likely there's a much simpler condition than that, since according to Wikipedia it's not even known when there's a solution to $x^2 - Dy^2 = -1$.
answered Sep 19, 2023 at 16:23
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2$\begingroup$ It is exactly known when there's a solution to $x^2 - Dy^2 = -1$. What is not known is a simple elementary short answer. $\endgroup$ Commented Sep 20, 2023 at 14:55