Pell Equations and Generalized Pell Orbits — Examples

A direct continued-fraction calculation gives $\sqrt{2}=[1;\bar{2}]$, so the period length is $\mathrm{\ell }=1$. Fact [F1] therefore gives the least positive negative-Pell solution $p_0\mathrm{/}{q}_0=1\mathrm{/}1,$ hence $1+\sqrt{2}$, and the least positive norm-one solution $p_1\mathrm{/}{q}_1=3\mathrm{/}2,$ hence ${\epsilon }_2=3+2\sqrt{2}=(1+\sqrt{2}{)}^2$.

By [F2], every positive Pell solution for $D=2$ is a power of ${\epsilon }_2$. Multiplying out the first powers gives ${\epsilon }_2^2=(3+2\sqrt{2}{)}^2=17+12\sqrt{2},{\epsilon }_2^3=(17+12\sqrt{2})(3+2\sqrt{2})=99+70\sqrt{2}\mathrm{.}$

A direct continued-fraction calculation gives $\sqrt{3}=[1;\overline{1,2}]$, so the period length is $\mathrm{\ell }=2$. Then [F1] gives the least positive norm-one solution at $p_1\mathrm{/}{q}_1=2\mathrm{/}1,$ hence $2+\sqrt{3}$, with $2^2-3\cdot 1^2=1.$

Because the period length is even, [F1] also says that $x^2-3y^2=-1$ has no integral solution.

A direct continued-fraction calculation gives $\sqrt{5}=[2;\bar{4}]$, so $\mathrm{\ell }=1$. Fact [F1] therefore gives the least positive negative-Pell solution $p_0\mathrm{/}{q}_0=2\mathrm{/}1,$ that is, $2+\sqrt{5}$, and indeed $2^2-5\cdot 1^2=-1.$

Squaring the negative-Pell solution yields $(2+\sqrt{5}{)}^2=9+4\sqrt{5},$ and $9^2-5\cdot 4^2=1.$ By [F1], this is the least positive norm-one solution.

Starting from $\sqrt{14}$, rationalizing reciprocals gives the state cycle ${\alpha }_1=\frac{\sqrt{14}+3}{5},{\alpha }_2=\frac{\sqrt{14}+2}{2},{\alpha }_3=\frac{\sqrt{14}+2}{5},{\alpha }_4=\sqrt{14}+3,$ so the digits are $1,2,1,6$ and $\sqrt{14}=[3;\overline{1,2,1,6}]\mathrm{.}$ The convergents are therefore $3,4,\frac{11}{3},\frac{15}{4}\mathrm{.}$

Here $\mathrm{\ell }=4$ and $Q_4=1$, in agreement with [F1]. Applying [F2] at $n=3$ gives ${15}^2-14\cdot 4^2=(-1{)}^4{Q}_4=1,$ so $15+4\sqrt{14}$ is the least positive norm-one solution.

The displayed continued fraction has odd period length $\mathrm{\ell }=11$. Running the convergent recurrence through one full period gives $p_{10}\mathrm{/}{q}_{10}=29718\mathrm{/}3805.$ Fact [F1] therefore gives ${29718}^2-61\cdot {3805}^2=-1,$ so $29718+3805\sqrt{61}$ is the least positive negative-Pell solution.

Squaring that element gives $(29718+3805\sqrt{61}{)}^2=1766319049+226153980\sqrt{61},$ and therefore ${1766319049}^2-61\cdot {226153980}^2=1.$ By [F1], this is the least positive solution of Pell's equation for $D=61$. The example shows that a finite continued-fraction algorithm can still produce a very large fundamental solution.

The element $5+2\sqrt{6}$ has norm $1$, and there is no positive norm-one solution with $y=1$ because $x^2=7$ is impossible; hence $5+2\sqrt{6}$ is the fundamental Pell unit, so [F1] identifies the Pell powers with $(5+2\sqrt{6}{)}^k$. The element $3+\sqrt{6}$ has norm $3$. For $N=3$ and ${\epsilon }_6=5+2\sqrt{6}$, the bound of [F2] is $\mathrm{\mid }{x}_0\mathrm{\mid }\le \frac{\sqrt{3}}{2}(\sqrt{5+2\sqrt{6}}+(5+2\sqrt{6}{)}^{-1\mathrm{/}2})=3,$ $\mathrm{\mid }{y}_0\mathrm{\mid }\le \frac{\sqrt{3}}{2\sqrt{6}}(\sqrt{5+2\sqrt{6}}+(5+2\sqrt{6}{)}^{-1\mathrm{/}2})<2.$

The only integer solutions in that box are $(x,y)=(\pm 3,\pm 1)$. Moreover, $(3-\sqrt{6})=(3+\sqrt{6})(5-2\sqrt{6})=(3+\sqrt{6})(5+2\sqrt{6}{)}^{-1},$ so $(3,1)$ and $(3,-1)$ lie in the same Pell orbit, and multiplying by $-1$ gives the negative orbit. Therefore every solution of $x^2-6y^2=3$ is $\pm (3+\sqrt{6})(5+2\sqrt{6}{)}^k(k\in \mathbb{Z})\mathrm{.}$

The least positive norm-one solution for $D=7$ is $8+3\sqrt{7}$, since $8^2-7\cdot 3^2=1$ and the smaller positive values $y=1,2$ give $x^2=8,29$, neither a square. Substituting ${\epsilon }_7=8+3\sqrt{7}$ and $N=57$ into [F1] gives $\mathrm{\mid }{x}_0\mathrm{\mid }\le \frac{\sqrt{57}}{2}(\sqrt{8+3\sqrt{7}}+(8+3\sqrt{7}{)}^{-1\mathrm{/}2})<17,$ $\mathrm{\mid }{y}_0\mathrm{\mid }\le \frac{\sqrt{57}}{2\sqrt{7}}(\sqrt{8+3\sqrt{7}}+(8+3\sqrt{7}{)}^{-1\mathrm{/}2})<7.$

By [F2], it is enough to enumerate solutions in that finite box. The integer solutions there are exactly $(\pm 8,\pm 1)$ and $(\pm 13,\pm 4)$, because $8^2-7\cdot 1^2=57,{13}^2-7\cdot 4^2=57,$ and no other pair with $\mathrm{\mid }x\mathrm{\mid }\le 16$, $\mathrm{\mid }y\mathrm{\mid }\le 6$ satisfies the equation. To compare their Pell orbits, note that every nonzero power of ${\epsilon }_7=8+3\sqrt{7}$ has absolute value either at least ${\epsilon }_7$ or at most ${\epsilon }_7^{-1}$. But among the eight displayed solutions, every quotient of two distinct positive representatives has absolute value between $\frac{8-\sqrt{7}}{13+4\sqrt{7}}\text{and}\frac{13+4\sqrt{7}}{8-\sqrt{7}},$ and this interval lies strictly inside $({\epsilon }_7^{-1},{\epsilon }_7)$. So no ratio of two distinct displayed solutions is a power of ${\epsilon }_7$, and negative ratios are impossible because every power of ${\epsilon }_7$ is positive. Therefore these eight boxed solutions represent eight distinct Pell orbits.

One checks directly that $3^2-6\cdot 1^2=3,$ so $3+\sqrt{6}$ is a generalized Pell solution. A direct continued-fraction calculation gives $\sqrt{6}=[2;\overline{2,4}],$ whose convergents begin $2,\frac{5}{2},\frac{22}{9},\dots \mathrm{.}$ Thus $3\mathrm{/}1$ is not a convergent of $\sqrt{6}$.

The missing hypothesis is exactly the small-error bound in [F1]: here $\mid \sqrt{6}-\frac{3}{1}\mid =3-\sqrt{6}>\frac{1}{2}=\frac{1}{2\cdot 1^2},$ so [F1] does not apply. Therefore generalized Pell solutions need not all be convergents.

The element $u=\frac{1+\sqrt{5}}{2}$ is a unit in $\mathbb{Z}\text{\hspace{0.17em}⁣}\left[\frac{1+\sqrt{5}}{2}\right]$ because $u(u-1)=\frac{1+\sqrt{5}}{2}\cdot \frac{\sqrt{5}-1}{2}=1.$

If $u$ lay in $\mathbb{Z}[\sqrt{5}]$, then [F1] would give integers $x,y$ with $\frac{1+\sqrt{5}}{2}=x+y\sqrt{5}\mathrm{.}$ Comparing coefficients of $1$ and $\sqrt{5}$ yields $x=\frac{1}{2}$ and $y=\frac{1}{2}$, impossible in $\mathbb{Z}$. So $u$ is a unit of the larger quadratic order that is not in $\mathbb{Z}[\sqrt{5}]$.