Solving x^4=a mod p, given a is a quadratic residue

Question

Given prime number $p\equiv 1 \pmod 4$. Prove if $a∈F_p^×$ is a quadratic residue then the congruence $$x^4 ≡ a \pmod p$$ has either no solutions or four solutions. Give examples of each case.

Answer 1

Let us put $a = b^2 \pmod p$. So $x^4 -b^2 = 0\mod p$, or $(x^2 - b)(x^2+b)=0\pmod p$. Since $p = 1\mod 4$, $-1$ is a quadratic residue (known theorem). Hence the previous equation can be written $(x^2-b)(x^2-\alpha^2 b) = 0\pmod p$ with $\alpha^2 = -1 \pmod p$. It is clear from this that if $b$ is a quadratic residue, say $b=c^2\pmod p$, the equation has four solutions $x = c, -c, \alpha c, -\alpha c$, otherwise it has no solution.