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][1]
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][2]

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


  [1]: http://math.berkeley.edu/~ianagol/farey.jpg
  [2]: http://math.berkeley.edu/~ianagol/pivotseq.jpg