I'm reading a book defining a "spread number" in a prime field a number $s \in \mathbb{F}_{p}$ verifying $s(1-s)$ being a square number.
Then, the book says without proof that:
- if $p = 1 (\mod 4)$ then there is $\frac{p+1}{2}$ spread numbers in $\mathbb{F}_{p}$.
- if $p = 3 (\mod 4)$ then there is $\frac{p+3}{2}$ spread numbers in $\mathbb{F}_{p}$.
I'm not really into algebra but I'd like to see how I can prove that.
Do you have any idea ?
Consider the conic $s(1-s)=y^2$ over $\mathbb F_p$. Now look at its projective closure, that is given by $C\colon sz-s^2=y^2$. This always has a rational point, and hence it is isomorphic to $\mathbb P^1$. It follows that $C$ has exactly $p+1$ points. Now let's look at the points at infinity, namely let's set $z=0$, getting $-s^2=y^2$. If $p=3\bmod 4$ this has no solutions (remember $(0,0)$ is not allowed). If $p=1\bmod 4$ then there are two solutions, yielding the two points $(s:y:z)=(1:\pm i:0)$. Now the number of "spread numbers" is the number of distinct $s$-coordinates of affine points on $C$. If $p=3\bmod 4$, all points are affine. Let $P=(s_0,y_0)\in C$. If $y_0=0$, then $s_0=0,1$. If $y_0\neq 0$, then the points $(s_0,y_0)$ and $(s_0,-y_0)$ both lie on $C$. Hence, the $p-1$ points with $y_0\neq 0$ can be partitioned in $(p-1)/2$ pairs of points with the same $s$-coordinate. It follows that the number of spread numbers is $(p-1)/2+2=(p+3)/2$. The case $p=1\bmod 4$ is analogous.