quadratic forms

**mancillaj3** · April 10th 2009, 10:51 PM

$x^{2} \cong -1(mod\ 122)$

**Soroban** · April 11th 2009, 03:13 AM

Hello, mancillaj3!

$x^2 \:\equiv\: -1 \!\! \pmod{122}$

Since $-1 \equiv 121 \!\!\pmod{122}$

. . we have: . $x^2 \:\equiv\:121\!\!\pmod{122}$

Therefore: . $x \:\equiv\:\pm11\!\!\pmod{122}$

**mancillaj3** · April 11th 2009, 09:42 AM

Originally Posted by Soroban

Hello, mancillaj3!

Since $-1 \equiv 121 \!\!\pmod{122}$

. . we have: . $x^2 \:\equiv\:121\!\!\pmod{122}$

Therefore: . $x \:\equiv\:\pm11\!\!\pmod{122}$

that means it has two solutions? but what about using legendre would it work $x^{2} \cong -1(mod\ 2*61)$

**ThePerfectHacker** · April 11th 2009, 04:17 PM

Originally Posted by mancillaj3

$x^{2} \cong -1(mod\ 122)$

You have $x^2\equiv 11^2(\bmod 122)$ therefore $x^2\equiv 11^2(\bmod 2)$ and $x^2\equiv 11^2(\bmod 61)$ . Therefore, $x \equiv \pm 11 (\bmod 2)$ and $x\equiv \pm 11(\bmod 61)$ . Now use Chinese remainder theorem to get solutions mod $122$ .

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