any nonzero integer is quadratic residue

Theorem. For every nonzero integer $a$ there exists an odd prime number $p$ such that $a$ is a quadratic residue modulo $p$ .

Proof. $1^{\\circ}.$ $a=2$ . We see that $3^{2}\\equiv 2\\pmod{7}$ and $7\mid 2$ , whence 2 is a quadratic residue modulo $7$ .
$2^{\\circ}.$ $2\\mid a$ but $a\eq 2$ . The number $1^{2}-a=1-a$ (which is odd and $\eq\\pm 1$ ) has an odd prime factor $p$ which does not divide $a$ . Thus $a$ is a quadratic residue modulo $p$ .
$3^{\\circ}.$ $a=3$ . We state that $4^{2}-3=13\\equiv 0\\pmod{13}$ and $13\mid 3$ . Therefore 3 is a quadratic residue modulo 13.
$4^{\\circ}.$ $a=5$ . We see that $4^{2}-5=11\\equiv 0\\pmod{11}$ and $11\mid 5$ , i.e. 5 is a quadratic residue modulo 11.
$5^{\\circ}.$ $2\mid a$ but $a\eq 3$ , $a\eq 5$ . Now the number $2^{2}-a=4-a$ (which is odd and $\eq\\pm 1$ ) has an odd prime factor $p$ . Moreover, $p\mid a$ since $p\mid 4$ . Accordingly, $a$ is a quadratic residue modulo $p$ .

单词	AnyNonzeroIntegerIsQuadraticResidue
释义	any nonzero integer is quadratic residue Theorem. For every nonzero integer $a$ there exists an odd prime number $p$ such that $a$ is a quadratic residue modulo $p$ . Proof. $1^{\\circ}.$ $a=2$ . We see that $3^{2}\\equiv 2\\pmod{7}$ and $7\mid 2$ , whence 2 is a quadratic residue modulo $7$ . $2^{\\circ}.$ $2\\mid a$ but $a\eq 2$ . The number $1^{2}-a=1-a$ (which is odd and $\eq\\pm 1$ ) has an odd prime factor $p$ which does not divide $a$ . Thus $a$ is a quadratic residue modulo $p$ . $3^{\\circ}.$ $a=3$ . We state that $4^{2}-3=13\\equiv 0\\pmod{13}$ and $13\mid 3$ . Therefore 3 is a quadratic residue modulo 13. $4^{\\circ}.$ $a=5$ . We see that $4^{2}-5=11\\equiv 0\\pmod{11}$ and $11\mid 5$ , i.e. 5 is a quadratic residue modulo 11. $5^{\\circ}.$ $2\mid a$ but $a\eq 3$ , $a\eq 5$ . Now the number $2^{2}-a=4-a$ (which is odd and $\eq\\pm 1$ ) has an odd prime factor $p$ . Moreover, $p\mid a$ since $p\mid 4$ . Accordingly, $a$ is a quadratic residue modulo $p$ .
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