Number of Solutions of the Hyperbola Equations over Finite Fields I have a problem with proving the number of points of the hyperbola equation $H_u: x^2 - y^2 = u$ (for every u > 0 in finite field $F_p$) in the finite fields. I have to prove that the number of solutions of the hyperbola equation $H_1: x^2 - y^2 = 1$ is the same as the number of solutions of the equation $H_u$: $x^2 - y^2 = u$ in every finite fields $F_p$, so $|H_1| = |H_u|$.
For example:
Equation $x^2 - y^2 = 1$ has 12 solutions in $F_{13}$ and $x^2 - y^2 = 7$ has 12 in $F_{13}$
Equation $x^2 - y^2 = 1$ has 16 solutions in $F_{17}$ and $x^2 - y^2 = 5$ has 12 in $F_{17}$
I have been able to prove that this is true if u is a quadratic residue, but I have not been able to prove that if u is a quadratic non-residue.
Picture of my proof of the number of solutions of the hyperbola equation if a is quadratic residue)
Can anyone please help me formulate proof by mapping if u is a quadratic non-residue for every $u > 0$ in finite field $F_p$ (where p is prime number)? It's also necessary to find the appropriate representation, as I found in the case of quadratic residues, as shown in the figure in the appendix.
For example with quadratic residues:
In finite field $F_{11}: x^2 - y^2 = u$: for u = 1 -> u = 5 ($\alpha = 4$)
(1,0) -> ($4 \cdot 1$, $4\cdot 0$)
(2,5) -> ($4 \cdot 2$, $4\cdot 5$)
(2, 6) -> ($4 \cdot 2$, $4 \cdot 6$)
(4, 2) -> ($4 \cdot 4$, $4 \cdot 2$)
(4, 9) -> ($4 \cdot 4$, $4\cdot 9$)
(7, 2) -> ($4 \cdot 7$, $4\cdot 2$)
(7, 9) -> ($4 \cdot 7$, $4\cdot 9$)
(9, 5) -> ($4 \cdot 9$, $4\cdot 5$)
(9, 6) -> ($4 \cdot 9$, $4\cdot 6$)
(1, 0) -> ($4 \cdot 1$, $4\cdot 0$)
So it means, that for $(x, y) \in H_1: \varphi(x, y) \rightarrow (4x, 4y)$ where $(4x, 4y) \in H_5$ because 5 is quadratic residue. I need to find something like this for quadratic non-residue, for example u = 6.
Thanks for help
 A: It is still not entirely clear to me what kind of a mapping you are looking for. The following will work irrespective of whether $u\neq0$ is a square in $\Bbb{F}_p$ or not:
$$\phi(x,y)=(\frac12[(u+1)x-(u-1)y],\frac12[(u+1)y-(u-1)x]).$$
So here $(x',y')=\phi(x,y)$ satisfies $x'^2-y'^2=u$ if and only if $x^2-y^2=1$, which was the goal as I understood it.

This formula is by no means the only possible one. I arrived at it by the following process, the elements of which were explained in the later comments after the circle/hyperbola mix-up was sorted out.

*

*The mapping $T:\Bbb{F}_p^2\to\Bbb{F}_p^2$, $$T(x,y)=(x+y,x-y)$$ is a bijection when $p>2$.

*It has an inverse $T^{-1}:\Bbb{F}_p^2\to\Bbb{F}_p^2$ given by $$T^{-1}(v,w)=(\frac12(v+w),\frac12(v-w)).$$

*The transformation $T$ is relevant here as $x^2-y^2=(x+y)(x-y)$. Meaning that $(v,w)=T(x,y)$ is a solution of $vw=z$ if and only if $(x,y)$ is a solution of $x^2-y^2=z$ (here $z$ is an arbitrary non-zero constant in $\Bbb{F}_p$).

*Then we have the mapping $S:\Bbb{F}_p^2\to\Bbb{F}_p^2$, $$S(v,w)=(v,uw).$$ Its relevance comes from the fact that $(v,w)$ is a solution of $vw=1$ if and only if $S(v,w)=(a,b)$ is a solution of $ab=u$.

My transformation $\phi$ is the composition
$$\phi=T^{-1}\circ S\circ T,$$
and it is hopefully obvious why it works.

It may be worth observing that when $u=1$ the mapping $\phi$ is the identity, and when $u=-1$ we get $\phi(x,y)=(y,x)$.

Answering the title question is trivial via the invertible transformation $T$. For all $u\in\Bbb{F}_p$, $u\neq0$, the equation $x^2-y^2=u$ has $p-1$ solutions $(x,y)\in\Bbb{F}_p^2$ because the equation $vw=u$ obviusly has $p-1$ solutions $(v,u/v)$, where $v$ is ranging over the non-zero elements of $\Bbb{F}_p$.
