Solvability of a particular Negative Pell's Equation

Question

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?

Answer 1

Clearly $\varepsilon = \frac{k + \sqrt{k^2-4}}2$ is a unit with norm $1$. If the fundamental unit has norm $-1$, then $\varepsilon$ must be a square. It is not difficult to show (!) that it must be the square of an element of the form $\frac{a + \sqrt{k^2-4}}2$, which is only possible if $a = 1$ and $k = 3$.

Answer 2

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 ($k=3$).