That notation looks ugly, I will use $\mathbb{F}_p$ for the field with $p$ elements Lets work completely in $\mathbb{F}_p$. We look for the solutions $(x,y)$ of $x^2 = y^2+1$. One is given by $P=(1,0)$ and we use the method of secants to find all: We intersect lines $y=\lambda (x-1)$ through $P$ with slope $\lambda$ with the hyperbola $x^2-y^2=1$. This gives $x^2- (\lambda(x-1))^2=1 \iff (x-1)(x+1) - \lambda^2(x-1)^2 =0$, having $x=1$ (giving $y=0$, so resulting in the point $P$) as solution. The second solution is that of $(x+1) - \lambda^2(x-1) =0 \iff x=- \frac{1+\lambda^2}{1-\lambda^2}$. To $\lambda = \pm 1$, there corresponds no solution, but all other $\lambda$ give us some $x$ and thus also some $y$ in an unique way. If $\lambda, \lambda^\prime$ give the same $(x,y)$, then $\lambda (x-1) = y = \lambda^\prime (x-1)$, giving $\lambda=\lambda^\prime$ if $x \neq 1$. If $x=1$, we have $1=x=- \frac{1+\lambda^2}{1-\lambda^2} \iff 2=0$, contradiction. Conversely, every $(x,y) \neq P=(1,0)$ on the hyperbola gives us $\lambda = \frac{y}{x-1}$ by the same arguments. We see that this gives us a bijection between the slopes $\lambda \neq \pm 1$ and the points $\neq P$ on the hyperbola- Thus the set of points $(x,y)$ on the hyperbola equals $1$ (for $P=(1,0)$) plus the number of $\lambda \neq \pm 1$, thus in total equals $p-1$. If $y=0$, then $x^2=1 \iff x= \pm 1$, thus there are exactly two points with $y=0$. Similar, $x=0$ yields $y^2=-1$, having only (two) solutions if $p \equiv 1 \mod 4$. If $x,y \neq 0$, there are always exactly four points with the same value of $x^2$, namely $(\pm x , \pm y)$. It's also possible to get this result from http://www.problem-solving.be/pen/viewtopic.php?t=134 . Counting again, we get $\frac{p-1-4}{4}+1+1=\frac{p+3}{4}$ values for $x^2$ if $p \equiv 1 \mod 4$ and $\frac{p-1-2}{4}+1=\frac{p+1}{4}$ such values for $p \equiv -1 \mod 4$.