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Determine all $ N, |N| \le 9 $ such that $ x^2-94y^2=N $ is solvable in positive integers.

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Thanks, Mariano. I was thinking of the formatting on another site. – James Nov 1 '10 at 8:27
up vote 6 down vote accepted

You are asking for which $N$ there is an element of norm $N$ in the quadratic ring $R=\mathbb{Z}[\sqrt{94}]$. Now $R$ is the full ring of integers in the field $\mathbb{Q}(\sqrt{94})$ which has class number one. The problem is almost the same as determining for which $N$ there is a norm $N$ ideal in $R$. There is a norm $N$ ideal in $R$ iff either $N$ or $-N$ is a norm from $R$. But we can't have both $N$ and $-$ being norms, since $-1$ is not a norm from $K$.

So the question reduces to (i) for which $N$ is there an ideal in $R$ of norm $N$? (ii) if there is such an ideal, is the corresponding solution to Pell's equation with $+N$ or $-N$?

For (i) $N=p_1^{r_1}\cdots p_k^{r_k}$ is a norm from $R$ iff $r_j$ is even whenever $p_j$ is inert in $R$. Now $p$ is inert iff the Legendre symbol $(94/p)=-1$.

For (ii) we can sometimes use congruence conditions. From (i) for instance $x^2-94y^2=\pm3$ is soluble, but thinking modulo $8$ we see that $x^2-94y^2=-3$ is insoluble. So $x^2-94y^2=3$ is soluble.

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Thank you Robin. – James Nov 2 '10 at 4:30

Putting the bound on $N$ at $9$ seems quite non-random. The continued fraction method for solving $x^2 - dy^2 = 1$ also works on $x^2 - dy^2 = N$ as long as $|N| < \sqrt{d}$, which corresponds exactly to your bound on $N$ when $d = 94$. Therefore your question almost sounds like it was set by someone who knew continued fractions. How did you come up with your question?

Generally, your questions would be better posed if you explain where they come from instead of writing them in the "Solve this" format, which makes them look suspiciously like homework tasks.

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