let p be a prime number such that {p = 3 [4]} show that the equation {x ^ 2 + 1} = {[p]} does not admit solutions in Z

Question

let p be a prime number such that {p = 3 [4]} show that the equation {x ^ 2 + 1} = {[p]} does not admit solutions in Z

Answer 1

Do you mean that $p \equiv 3 \pmod 4$, that is, $p$ is of the form $4k + 3$, where $k$ is an integer?

Then it suffices to note that in order for $x^2 + 1 = p$ to be true, $x^2$ must be even. If $x$ is an even integer, say $x = 2m$, where $m$ may be odd or even, then $x^2 = 4m^2$. But then $x^2 + 1 \equiv 1 \pmod 4$, not $3$.

In order for $x^2 = 4k + 2$ to be true, we need to solve $x = \sqrt{4k + 2}$. That would be $\sqrt 2 \sqrt{2k + 1}$. This is guaranteed not to be an integer even if $\sqrt{2k + 1}$ is.