<s>[INST] Solvability of a particular Negative Pell's Equation How to show that the equation $x^2-(k^2-4)y^2=-1$ is solvable only when k=3?
I can show that 3 must divide d otherwise 3 divides $k^2-4$ and the equation will not be solvable.Again k=3(2m+1) because other wise $x^2\equiv -1(mod8)$ which is not possible.Now how should I show that m=0?
 [/INST] Another approach:
Consider equation $x^2-Dy=-1$
Let $D=m^2+1$. we have:
$x^2-m^2y^2-y^2=-1$
$\Rightarrow (x-y)(x+y)=(my-1)(my+1)$
Suppose:
$\begin{cases} x-y=my-1\\x+y=my+1\end{cases}$
which gives :
($x=my$) and ($y=1$) and we have:
$D=m^2-1=k^2-4\Rightarrow (m-k)(m+k)=-5$
This is possible only if ($m=2$) and ($k=3$).
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