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.
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Pick, for example, $p=11$. There is no $x$ such that $x^2-2x+4$ is divisible by $11$. – André Nicolas Sep 22 '14 at 12:23
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Seems to me like there's a missing clause, like an additional stricture on $p$. – Sep 22 '14 at 16:56
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HINT:
$$p\equiv3\pmod4\equiv-1$$
$$x^2-2x+4\equiv0\pmod p\iff(x-1)^2\equiv-3\pmod p$$
Check when $$\left(\frac{-3}p\right)=1$$
lab bhattacharjee
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Sir, please explain the last part. I am slow learner. Please understand me. If p =3, then p = 3 (mod 4). But you said p = 3 (mod 4) = -1 and so on... – user177889 Sep 22 '14 at 11:21
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@user177889, http://math.stackexchange.com/questions/881341/let-p-3-be-a-prime-number-show-that-x2-equiv-%E2%88%923-mod-p-is-solvable-iff?lq=1 – lab bhattacharjee Sep 22 '14 at 11:26
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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 $3^2 + 3, 6^2 + 3, 7^2 + 3, \ldots$