divisibility on prime and expression This site is amazing and got good answer.
This is my last one.
If $4|(p-3)$ for some prime $p$, then $p|(x^2-2x+4)$.
can you justify my statement?
High regards to one and all.

If $x \geq 0$, then $x^2 - 2x + 4 = (x - 1)^2 + 3$.
So then you want to see if $-3$ is a quadratic residue modulo $p$, and that's what you use the Legendre symbol $(\frac{-3}{p})$ for, which gives you a yes ($1$) or no ($-1$) answer.
But $p \equiv 3 \mod 4$ does not guarantee $(\frac{-3}{p}) = 1$, as André's example of $p = 11$ shows. And even if $(\frac{-3}{p})$ does equal $1$, all that tells you is that there is at least one integer $x$ such that $p|(x^2 + 3)$; for example, with $p = 7$, we readily see that $p$ is a divisor of $2^2 + 3, 5^2 + 3, 9^2 + 3, \ldots$ but not $3^2 + 3, 6^2 + 3, 7^2 + 3, \ldots$
