Negative Pell is soluble exactly for odd period length

Applying [F4] at $n=\mathrm{\ell }-1$ and using $Q_{\mathrm{\ell }}=1$ from [F1] gives $p_{\mathrm{\ell }-1}^2-D{q}_{\mathrm{\ell }-1}^2=(-1{)}^{\mathrm{\ell }}\mathrm{.}$ Hence $\mathrm{\ell }$ odd yields a negative-Pell solution at $p_{\mathrm{\ell }-1}\mathrm{/}{q}_{\mathrm{\ell }-1}$, while $\mathrm{\ell }$ even yields a positive norm-one solution there.

Conversely, suppose $x,y>0$ satisfy $x^2-D{y}^2=-1$. Then $0<\sqrt{D}-\frac{x}{y}=\frac{1}{y(x+y\sqrt{D})}\mathrm{.}$ Because $D\ge 2$, one has $x^2=D{y}^2-1\ge y^2$, so $x\ge y$, and therefore $x+y\sqrt{D}>2y\mathrm{.}$ Hence $\mid \sqrt{D}-\frac{x}{y}\mid <\frac{1}{2y^2}\mathrm{.}$ Since $\gcd (x,y)=1$ follows from $x^2-D{y}^2=-1$, fact [F5] shows that $x\mathrm{/}y=p_n\mathrm{/}{q}_n$ for some convergent.

By [F1] and [F3], the returned state at one full period is ${\alpha }_{\mathrm{\ell }}=\frac{\sqrt{D}+P_{\mathrm{\ell }}}{Q_{\mathrm{\ell }}}=\sqrt{D}+a_0\mathrm{.}$ Since $\lfloor {\alpha }_{\mathrm{\ell }}\rfloor =2a_0$, fact [F2] gives ${\alpha }_{\mathrm{\ell }+1}=\frac{1}{{\alpha }_{\mathrm{\ell }}-\lfloor {\alpha }_{\mathrm{\ell }}\rfloor }=\frac{1}{\sqrt{D}-a_0}={\alpha }_1\mathrm{.}$ Thus one full period returns the complete-quotient algorithm to ${\alpha }_1$; because the digits from $a_1$ onward repeat with period $\mathrm{\ell }$, the same computation one period later gives ${\alpha }_{2\mathrm{\ell }}=\sqrt{D}+a_0$, hence $Q_{2\mathrm{\ell }}=1$. Applying [F4] at $n=2\mathrm{\ell }-1$ yields $p_{2\mathrm{\ell }-1}^2-D{q}_{2\mathrm{\ell }-1}^2=(-1{)}^{2\mathrm{\ell }}{Q}_{2\mathrm{\ell }}=1.$ Thus $p_{2\mathrm{\ell }-1}\mathrm{/}{q}_{2\mathrm{\ell }-1}$ is a positive norm-one solution.

For that index $n$, [F4] gives $-1=x^2-D{y}^2=p_n^2-D{q}_n^2=(-1{)}^{n+1}{Q}_{n+1}\mathrm{.}$ Since $Q_{n+1}$ is a positive integer, it follows that $Q_{n+1}=1\text{and}n\text{ is even}\mathrm{.}$ Facts [F2] and [F3] then give ${\alpha }_{n+1}=\sqrt{D}+P_{n+1},{\alpha }_{n+2}=\frac{1}{{\alpha }_{n+1}-\lfloor {\alpha }_{n+1}\rfloor }=\frac{1}{\sqrt{D}-a_0}={\alpha }_1\mathrm{.}$ So the digit block from $a_1$ onward repeats with period $n+1$. Because $\mathrm{\ell }$ is the period length, $\mathrm{\ell }$ divides $n+1$. Since $n+1$ is odd, $\mathrm{\ell }$ must be odd.

Step 2.1 shows that every positive negative-Pell solution comes from a convergent index $n$ with $n$ even and $\mathrm{\ell }\mid (n+1)$, so the first such index is $\mathrm{\ell }-1$. Now let $u,v>0$ satisfy $u^2-D{v}^2=1$. Then $0<\frac{u}{v}-\sqrt{D}=\frac{1}{v(u+v\sqrt{D})}\mathrm{.}$ Because $D\ge 2$, one has $u^2=D{v}^2+1>v^2$, so $u>v$, and therefore $u+v\sqrt{D}>2v\mathrm{.}$ Hence $\mid \sqrt{D}-\frac{u}{v}\mid <\frac{1}{2v^2}\mathrm{.}$ Since $\gcd (u,v)=1$, [F5] shows that $u\mathrm{/}v=p_n\mathrm{/}{q}_n$ for some convergent. Applying [F4] gives $1=u^2-D{v}^2=p_n^2-D{q}_n^2=(-1{)}^{n+1}{Q}_{n+1},$ so $Q_{n+1}=1\text{and}n\text{ is odd}\mathrm{.}$ As in step 2.1, facts [F2] and [F3] then give ${\alpha }_{n+2}={\alpha }_1$, so $\mathrm{\ell }$ divides $n+1$. Therefore the first possible norm-one index is $\mathrm{\ell }-1$ when $\mathrm{\ell }$ is even and $2\mathrm{\ell }-1$ when $\mathrm{\ell }$ is odd. Convergent denominators strictly increase, and for positive solutions of $x^2-D{y}^2=\pm 1$ the value of $x$ is determined by $x^2=D{y}^2\pm 1$, so the first convergent of each type gives the least positive solution. Therefore, when $\mathrm{\ell }$ is odd, $p_{\mathrm{\ell }-1}\mathrm{/}{q}_{\mathrm{\ell }-1}$ is the least positive negative-Pell solution and $p_{2\mathrm{\ell }-1}\mathrm{/}{q}_{2\mathrm{\ell }-1}$ is the least positive norm-one solution, while when $\mathrm{\ell }$ is even, $p_{\mathrm{\ell }-1}\mathrm{/}{q}_{\mathrm{\ell }-1}$ is the least positive norm-one solution.