Question about negative Pell's Equation

Question

Is it true that, if $a^2-Db^2=-1$ is solvable in integers, then so is $x^2-Dy^2=D$ (*)?

For $D=5$ this is true, you can take $x=5$ and $y=2$, and indeed $5^2-5(2^2)=5$, so (*) is solvable. Is this true in general?

Answer 1

Standard fact in Gaussian composition, or more precisely Dirichlet composition. If the principal form of a discriminant integrally represents $-1,$ then it is equivalent to its own negative, and every form of that discriminant represents some integer $k$ if and only if it represents $-k.$ You had the substitution correct in your other question, if $u^2 - n v^2 = -1,$ then $$ \left( \begin{array}{cc} u & v \\ nv & u \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & -n \end{array} \right) \left( \begin{array}{cc} u & nv \\ v & u \end{array} \right) = \left( \begin{array}{cc} -1 & 0 \\ 0 & n \end{array} \right) $$

It is true that the transformation matrix on the right, $$ \left( \begin{array}{cc} u & nv \\ v & u \end{array} \right) $$ has determinant $-1,$ but we can just negate one row or one column to get determinant $1$ and it still works. For example, $$ \left( \begin{array}{cc} u & v \\ -nv & -u \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & -n \end{array} \right) \left( \begin{array}{cc} u & -nv \\ v & -u \end{array} \right) = \left( \begin{array}{cc} -1 & 0 \\ 0 & n \end{array} \right) $$ and $$ \left( \begin{array}{cc} u & -nv \\ v & -u \end{array} \right) $$ has determinant $+1.$