I want to understand why the equation $U^2-(m^2-4)V^2=-4$ (when $U,V,m$ are odd number and $m > 3$) is impossible. (This came from a post I was reading here)
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1If there is an odd solution $(u,v)$ to $u^2-Dv^2=-4$, then $(x,y)=((u^2+3)u/2, (u^2+1)v/2)$ is a solution to $x^2-Dy^2=-1$. This equation has no solution for $D=m^2-4$, if $m>3$ is odd. – Dietrich Burde Aug 06 '13 at 21:02
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@DietrichBurde Thanks. May I ask how did you get $x$ and $y$ ? and why the new equation $x^2-Dy^2 = -1$ has no solution ? What is the theorem who says that !? – A A Aug 06 '13 at 21:11
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Let $D=m^2-4$ with $m\ge 5$ odd. If there is an odd solution $(u,v)$ to $u^2-Dv^2=-4$, then $(x,y)=(\frac{(u^2+3)u}{2}, (\frac{(u^2+1)v}{2})$ is a solution to $x^2-Dy^2=-1$. Hence it is enough to show that the latter negative Pell equation has no solution for $D=m^2-4$ with $m\ge 5$ odd. It is well known that $x^2-Dy^2=-1$ is solvable in integers $x,y$ if and only if $\ell(\sqrt{D})$ is odd. Here $\ell(\sqrt{D})$ is the length of the continued fraction of the quadratic irrational $\sqrt{D}$. But, with $m=n+1$, $n>3$ even, we have $$ \sqrt{D}=\sqrt{n^2+2n-3}=[n;\overline{1,\frac{n-2}{2},2,\frac{n-2}{2},1,2n}], $$ as an easy calculation shows. Hence $\ell(\sqrt{D})=6$ is even, and there is no integer solution. Note that for $n=2$ this is not true, and there is a solution of $x^2-5y^2=-1$, namely $(x,y)=(2,1)$. Then $U^2-5V^2=-4$ also has a solution, namely $(U,V)=(1,1)$.
Edit: This is a corrected answer, the comments refer to an earlier version, which was not complete.
Dietrich Burde
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This is exactly what I don't understand (and that why I proposed the problem above): How do you reduce problem to $U^2-DV^2 = -1$ (of course if $U$ and $V$ are even then we can divide by 4 but what if they are both odd) ?? – A A Aug 06 '13 at 19:51
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