If $p$ is a prime of the form $4k+3$ and $n$ is any integer, then $p$ doesn't divide $n^2+1$.

Question

If $p$ is a prime of the form $4k+3$ and $n$ is any integer, then $p$ doesn't divide $n^2+1$.

The "$p$ doesn't divide $n^2+1$" part is giving me real trouble when trying to prove this.

Any help or hints with the problem is appreciated.

Answer 1

Hint:

$$\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}$$

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Assuming you are to prove this without the Legendre symbol:

You are given that $p\not\mid n^2+1$, which can be formulated as $$n^2+1\not\equiv_p0\iff n^2\not\equiv_p -1\tag{1}$$ Now raise both sides of eq. $(1)$ to $\frac{p-1}{2}$, this yields $$n^{p-1}\not\equiv_p(-1)^{\frac{p-1}{2}}\tag{2}$$ By Fermat's Little Theorem we have $$1\not\equiv_p(-1)^{\frac{p-1}{2}}\tag{3}$$

What can you say about the expression $\frac{p-1}{2}$ then?