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I have solved the Pell equation $ p^2 - 95 q^2 =1$ . By looking at the convergents corresponding to the simple continued fraction of $\sqrt{95}$ I was able to find the fundamental solution $p=39$ and $q=4$ . I found the five smallest pairs of positive integers $p,q$ that satisfy the above Pell equation. They are : $$\begin{align*} p=39 \quad& q=4 \\ p=3041 \quad&q=312\\ p=237159 \quad& q=24332 \\ p=18495361\quad& q=1897584 \\ p=1442400999 \quad& q=147987220 \end{align*}$$

However I am having difficulty solving the related Pell equation $$ p^2 - 95 q^2 =-1 , +1 , -1 , +1 , -1 , +1 , .....$$

The only difference now is that the right hand side of the equation is alternatively $-1$ and $+1$ , instead of just $+1$. One obvious trivial solution is $p=1$ and $q=0$ , or $p= \sqrt{-1}$ and $q=0$ but these trivial solutions do not count.

I am trying to find the five smallest pairs of positive integers $ p,q$ that satisfy this equation. I would appreciate your help.

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    $\begingroup$ Take remainders $\mod 4$, you need $p^2+q^2\equiv3\pmod4$ $\endgroup$ – Empy2 Jul 17 '20 at 10:52
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There are no solutions to $p^2-95q^2=-1$, because they would imply $p^2\equiv-1\bmod95$, which would imply $p^2\equiv-1\bmod19$, and there are no solutions to $p^2\equiv-1\bmod19$, because there are no solutions to $p^2\equiv-1\bmod n$ for prime $n\equiv-1\bmod4.$

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I agree with J W Tanner's answer. This response is intended to provide alternative analysis.

Given $D \,\in \,\mathbb{Z^+},$ where $D$ is not a perfect square, let
$[a_0; \overline{a_1, a_2, \cdots, a_{n-1}, 2a_0}]$ represent the representation of
$\sqrt{D}$ as a (simple) continued fraction.

This means that the length of the period of this representation is $n.$

Then, it is well settled that if $n$ is even
then the Diophantine equation $x^2 - Dy^2 = -1$ will have no positive integer solutions.

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