Question

It is a known fact that for a Prime $P$, $P\equiv 1$ mod $4$ iff the length of the period in the repeating block for the continued fraction expansion of $\sqrt{P}$ is odd. I have an elementary proof of this using the classical result: $P\equiv 1$ mod $4$ iff $x^2-Py^2=-1$ has integer solutions and a proof that $x^2-Py^2=-1$ has integer solutions iff the length of the period in the repeating block for the continued fraction expansion of $\sqrt{P}$ is odd.

I have tried repeatedly to give a direct elementary proof that $P\equiv 1$ mod $4$ implies that the length of the period in the repeating block for the continued fraction expansion of $\sqrt{P}$ is odd, but cant seem to figure it out. (I have a direct elementary proof of the converse)

Does anyone know of an elementary proof of this result or where I may find one?

Answer 1

This doesn't answer your question, but it might gives some intuition as to why this is true. Given the fact that there exists a solution to the equation $x^2-Py^2=-1$ when $P\equiv 1(\mod 4)$, one sees that the matrix $$ \begin{pmatrix}x & Py \\ y & x\end{pmatrix} $$ fixes $\pm\sqrt{P}$ (under the action of $PGL_2(\mathbb{Z})$ on $\mathbb{RP}^1=\partial_{\infty} \mathbb{H}^2$). The conjugacy class of a primitive matrix in $GL_2(\mathbb{Z})$ (which is not reducible or finite order) is determined by the closed geodesic on the modular orbifold that it represents. This in turn is determined by a sequence of triangles which the geodesic crosses in the Farey graph: These triangles come in bunches sharing a common vertex, where the number in each bunch corresponds to coefficients of the continued fraction expansion. The matrix is conjugate to $$\pm \left[\begin{array}{cc}1 & a_1 \\ 0 & 1\end{array}\right] \left[\begin{array}{cc}1 & 0 \\ a_2 & 1\end{array}\right] \cdots \left[\begin{array}{cc}1 & a_{2n} \\ 0 & 1\end{array}\right]$$ if the determinant is 1, and to $$\pm \left[\begin{array}{cc}1 & a_1 \\ 0 & 1\end{array}\right] \left[\begin{array}{cc}1 & 0 \\ a_2 & 1\end{array}\right] \cdots \left[\begin{array}{cc}1 & 0 \\ a_{2n-1} & 1\end{array}\right] \left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right] $$ if the determinant is $-1$.

[Remark: the labels in this figure don't quite correspond to the matrices - it should be $a_i$'s instead of $\alpha_i$'s, and $\alpha_{\pm}$ should be $\pm\sqrt{P}$]

The number of such factors corresponds precisely to the period of the continued fraction expansion of fixed points of the matrix, since the closed geodesic is asymptotic in $\mathbb{H}^2$ to the geodesic connecting $\infty$ to $\sqrt{P}$, whose Farey sequence gives rise to the continued fraction expansion of $\sqrt{P}$. This number is even if and only if the matrix is orientation preserving, which is if and only if the determinant is 1. So the determinant is $1$ if and only if the continued fraction has even period, and the determinant is $-1$ if and only if the continued fraction has odd period, corresponding to $P\equiv 1(\mod 4)$.