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Let $m$ be a positive integer and $a$ any integer such that $(a, m) = 1$. Then $a$ is a quadratic residue of $m$ if the congruence $x^2 ≡ a (mod m)$ is solvable; otherwise, it is a quadratic nonresidue of $m$. For example if we want to find the quadratic residues of $m = 13$, we calculated $x^2 ≡ a (mod 13)$ with $x=1,2,3,4,5,6,7,8,9,10,11,12$ and we will find that the quadratic residues are $a = 1, 3, 4, 9, 10,12$. If we continue calculating for $x=14$, $x=15$ and so on, we notice that the residuals will be repeated $a=1$, $a=4$ and so on,I wanted to know why this happens? Why is it enough to calculate the squares of numbers $x=1,2,3,4,5,6,7,8,9,10,11,12$ only?

Ilovemath
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  • By the linked Congruence Product or Power Rule $, x\equiv \bar x\Rightarrow x^2\equiv \bar x^2.,$ More generally $, x\equiv \bar x\Rightarrow f(x)\equiv f(\bar x),$ for any polynomial with integer coef's, by the Polynomial Congruence Rule, so we need only test for roots in a complete residue system, since every integer $x$ is congruence to an integer $\bar x$ in the complete residue system. – Bill Dubuque Aug 27 '22 at 00:39
  • As an analogy, if we wish to test if a polynomial has a rational (fraction) root, then it suffices to test fractions in least terms, since every fraction is equivalent to one in least terms (like every integer $\bmod n$ is equivalent to its remainder $\bmod n,,$ i.e. the remainder = least positive rep. may be viewed as its "least terms" residue). See here for more. – Bill Dubuque Aug 27 '22 at 00:51

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