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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: alt text http://math.berkeley.edu/%7Eianagol/farey.jpg 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 $-1$. alt text http://math.berkeley.edu/%7Eianagol/pivotseq.jpg

[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 $P\equiv 1(\mod 4)$.

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 $P\equiv 1(\mod 4)$.

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 $$\left[\begin{matrix}x & Py \\ y & x\end{matrix}\right]$$ fixes $\pm\sqrt{P}$ (under the action of $PGL_2(\mathbb{Z})$ on $\mathbb{PR}^1=\partial_{\infty} \mathbb{H}^2$). The conjugacy class of a 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: alt text http://math.berkeley.edu/%7Eianagol/farey.jpg 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 $$\pm \left[\begin{array}{cc}1 & a_1 \\ 0 & 1\end{array}\right] \left[\begin{array}{cc}1 & 0 \\ 1 & a_2\end{array}\right] \cdots \left[\begin{array}{cc}1 & a_{2n} \\ 0 & 1\end{array}\right]$$ if the determinant is 1, and to $$\pm \left[\begin{array}{cc}1 & a_1 \\ 0 & 1\end{array}\right] \left[\begin{array}{cc}1 & 0 \\ 1 & a_2\end{array}\right] \cdots \left[\begin{array}{cc}1 & 0 \\ a_{2n-1} & 1\end{array}\right] \left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right] $$ if the determinant is $-1$. alt text http://math.berkeley.edu/%7Eianagol/pivotseq.jpg

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 $P\equiv 1(\mod 4)$.

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: alt text http://math.berkeley.edu/%7Eianagol/farey.jpg 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 $-1$. alt text http://math.berkeley.edu/%7Eianagol/pivotseq.jpg

[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 $P\equiv 1(\mod 4)$.

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 $x^2-Py^2=-1$, one sees that the matrix $$\left[\begin{matrix}x & Py \\ y & x\end{matrix}\right]$$ fixes $\pm\sqrt{P}$. The conjugacy class of a 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. alt text http://math.berkeley.edu/%7Eianagol/farey.jpg 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 $$\pm \left[\begin{array}{cc}1 & a_1 \\ 0 & 1\end{array}\right] \left[\begin{array}{cc}1 & 0 \\ 1 & a_2\end{array}\right] \cdots \left[\begin{array}{cc}1 & a_{2n} \\ 0 & 1\end{array}\right]$$ if the determinant is 1, and to $$\pm \left[\begin{array}{cc}1 & a_1 \\ 0 & 1\end{array}\right] \left[\begin{array}{cc}1 & 0 \\ 1 & a_2\end{array}\right] \cdots \left[\begin{array}{cc}1 & 0 \\ a_{2n-1} & 1\end{array}\right] \left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right] $$ if the determinant is $-1$. The number of such factors corresponds precisely to period 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 $\sqrt{P}$. The 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.

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 $$\left[\begin{matrix}x & Py \\ y & x\end{matrix}\right]$$ fixes $\pm\sqrt{P}$ (under the action of $PGL_2(\mathbb{Z})$ on $\mathbb{PR}^1=\partial_{\infty} \mathbb{H}^2$). The conjugacy class of a 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: alt text http://math.berkeley.edu/%7Eianagol/farey.jpg 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 $$\pm \left[\begin{array}{cc}1 & a_1 \\ 0 & 1\end{array}\right] \left[\begin{array}{cc}1 & 0 \\ 1 & a_2\end{array}\right] \cdots \left[\begin{array}{cc}1 & a_{2n} \\ 0 & 1\end{array}\right]$$ if the determinant is 1, and to $$\pm \left[\begin{array}{cc}1 & a_1 \\ 0 & 1\end{array}\right] \left[\begin{array}{cc}1 & 0 \\ 1 & a_2\end{array}\right] \cdots \left[\begin{array}{cc}1 & 0 \\ a_{2n-1} & 1\end{array}\right] \left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right] $$ if the determinant is $-1$. alt text http://math.berkeley.edu/%7Eianagol/pivotseq.jpg

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 $P\equiv 1(\mod 4)$.