Show $x^2+y^2 \equiv1\pmod p$ has $p-1$ solutions if $p \equiv1\pmod4$ and...

Question

Question:

Show the equation $x^2+y^2 \equiv1\pmod p$ has $p-1$ solutions if $p \equiv1\pmod4$, and $p+1$ solutions if $p \equiv 3\pmod4$

I'm really stuck on this one. Any help would be highly appreciated.

Answer 1

If $p$ is a prime of the form $4k+1$,

  Let $r$ be a primitive root mod $p$ (which can be proven by various means)

  Let $i = r^k$

  $i^2 = r^{2k} = -1$ (mod $p$) because $(i^2)^2 = r^{p-1} = 1$ (mod $p$)

  If $1 = x^2-(iy)^2 = (x+iy)(x-iy)$,

    Let $a = x+iy$

    $x = (a+a^{-1})2^{-1}$

    $y = (a-a^{-1})(2i)^{-1}$

  Therefore there is a bijection between $\{ (x,y) : x^2+y^2 = 1 \space(\text{mod } p)\}$ and $\{ a : a \in [1..p-1] \}$

If $p$ is a prime of the form $4k+3$,

  I can't prove it.