Number of solution of $x^2+1 \equiv 0 \pmod p$ where $p\equiv 1 \pmod 4$ and p is a prime

elementary-number-theoryquadratic-residues

This question is to be used in a theorem of Algebraic Number Theory and I am struck on this.

Question: Prove that there are 2 solutions of the equation $x^2 +1 \equiv 0 \pmod p$ . Here $p\equiv 1 \pmod 4$ and p is a prime.

By Euler's criteria, −1 is quadratic residue. So, solution of the equation exists mod $p$. But I am not able to prove that only two solutions exists.

Can you please tell which result should I use?

I have studied elementary number theory from David Burton.

Thanks!

Best Answer

$\mathbb{Z}_p$ is a field so, $x^2+1 $ can have atmost 2 roots. Atleast 1 root exists by Euler's Criterion of Quadratic Residues and it x is given a solution, then -x is also a solution and if x=-x , then this implies that 2x=p or 2|p but p is prime , so $x\neq -x$. So, exactly 2 roots exists.

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