If $p\equiv3 \pmod4 $, then there does not exist any integer $x$ with $x^2 ≡ −1 \pmod p$.

Question

Suppose that $p$ is a prime number that that $p\equiv3 \pmod4 $. Prove that there does not exist any integer $x$ with $x^2 ≡ −1 \pmod p$.

I am looking to use Fermat's Little Theorem but not able to proceed in correct direction. Please give some hint.

Answer 1

Using Fermat's Little Theorem one has $x^{p-1}\equiv 1\pmod{p}$.

Now $p=4k +3$ leads to $\frac{p-1}2=2k+1\equiv 1\pmod{2}$.

So if we assume $x^{2}\equiv -1\pmod{p}$ we get

$$x^{p-1}=\left(x^2\right)^{{p-1\over 2}}\equiv -1\pmod{p}$$

In contradiction with Fermat's Little Theorem.

Answer 2

Hint:

Fermat's Little Theorem tells you that $x^{p-1} \equiv 1 \pmod p$.

What do you know about $\frac{p-1}{2}$ because of $p \equiv 3 \pmod 4$?

What do you know about $(x^2)^{\frac{p-1}{2}} \pmod p$ if $x^2 \equiv -1 \pmod p$?