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Math Help - quadratic forms

  1. #1
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    quadratic forms

    x^{2} \cong -1(mod\  122)
    Last edited by mancillaj3; April 10th 2009 at 11:20 PM.
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  2. #2
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    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}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    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)
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  4. #4
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    Quote Originally Posted by mancillaj3 View Post
    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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