Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 11H / a

For the continued-fraction assertion, take the standard Pell equation hypothesis $N>0$ nonsquare. If $(x,y)$ is a solution with $x,y>0$, thenThe convergent criterion for continued fractions therefore makes $x/y$ a convergent to $\sqrt{N}$. Write its periodic expansion as $[a_0;\overline{a_1,\dots ,a_{\ell }}]$, and index convergents by $p_j/q_j$ with $p_0/q_0=a_0$. The norm-one convergents occur at $j=k\ell -1$ with $k\ell$ even. In particular the fundamental positive solution comes at $\ell -1$ if $\ell$ is even, and at $2\ell -1$ if it is odd. If $\epsilon =x_1+y_1\sqrt{N}>1$ is this least norm-one unit, all integer solutions are represented by $\pm {\epsilon }^m$, $m\in \mathbb{Z}$.

The abelian group law comes from multiplication of the norm-one expressions:Their field norms multiply, so the result remains a solution. Multiplication also proves associativity and commutativity; the identity is $(1,0)$ and the inverse of $(x,y)$ is $(x,-y)$. Cubing givesThe last form uses $N{y}^2=x^2-1$.

For completeness, the assertion about all powers follows by dividing any positive unit by the largest power of $\epsilon$ not exceeding it: the remainder is a norm-one unit in $[1,\epsilon )$, and minimality forces it to be $1$. Sign changes and inverses give the remaining solutions. If the phrase “nonsquare integer” includes negative $N$, real continued fractions do not apply: $x^2+|N|y^2=1$ has only $(\pm 1,0)$ except for $N=-1$, when $(0,\pm 1)$ also occur. The same multiplication law still works in those finite groups.