For any prime $p$ of the form $4k + 3$, prove that $x^2 + \frac{p+1}{4} \equiv 0 \;(\text{mod }p)$ is not solvable

Question

I tried using quadratic reciprocity to simplify $$\left(\frac{-\frac{p+1}{4}}{p}\right)$$ and tried replacing $p$ by $k$ $(\text{as in}\, p = 4k + 3)$, but don't know how to proceed after that.

Answer 1

$x^2+\frac{p-1}{4}\equiv 0\pmod{p}$ is solvable iff $\left(\frac{\color{red}{-}\frac{p+1}{4}}{p}\right) = 1$, that implies $\left(\frac{-1}{p}\right)=1$ by the multiplicativity of Legendre symbol. On the other hand $p\equiv 3\pmod{4}$ implies $\left(\frac{-1}{p}\right)=-1$.

Answer 2

If you multiply both sides by $4$, as @GregMartin suggested you get $$4x^2 + p + 1 \equiv 0\mod{p}.$$ This simplifies to $$4x^2 \equiv -1\mod{p},\text{ or }(2x)^2 \equiv -1\mod{p}.$$ But $-1$ is not a square $\mod{p}$.