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Math Help - solving cubic congruences

  1. #1
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    solving cubic congruences

    Find all the solutions of x^3+3x-8\equiv 0mod(33)

    I can only see
    x^3+3x-8\equiv 0 mod(3) and x^3+3x-8 \equiv 0 mod(11)

    however i cannot seem to condense these further and mod 11 by inspection does seem right
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  2. #2
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    Quote Originally Posted by FGT12 View Post
    Find all the solutions of x^3+3x-8\equiv 0mod(33)

    I can only see
    x^3+3x-8\equiv 0 mod(3) and x^3+3x-8 \equiv 0 mod(11)

    however i cannot seem to condense these further and mod 11 by inspection does seem right
    well i'll be the one who brings up thoughtless brute force, someone else could discuss more intelligently.

    easiest to just show you a screenshot from pari/gp. as you can see i solved it in two ways and the answers match.
    Attached Thumbnails Attached Thumbnails solving cubic congruences-cubiccong.png  
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  3. #3
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    Quote Originally Posted by FGT12 View Post
    Find all the solutions of x^3+3x-8\equiv 0mod(33)

    I can only see
    x^3+3x-8\equiv 0 mod(3) and x^3+3x-8 \equiv 0 mod(11)

    however i cannot seem to condense these further and mod 11 by inspection does seem right
    This is my attempt.
    For the first congruence: x^3\equiv x\bmod3 by Fermat's Theorem, and replacing coefficients modulo 3 gives x^3+3x-8\equiv x-8\equiv x+1\bmod3 or x\equiv-1\bmod3.
    Similarly, we replace coefficients modulo 11 and hence x^3+3x-8\equiv 12x^3+3x+3\equiv0\bmod11. Dividing by 3 gives 4x^3+x+1\equiv0\bmod11. Here by simply checking integers 0\le x\le10 gives x\equiv5, 8, 9\bmod11.

    Now we have to consider three systems: x\equiv-1\bmod 3 and x\equiv5\bmod11; x\equiv-1\bmod3 and x\equiv8\bmod11; x\equiv-1\bmod3 and x\equiv9\bmod11. By inspection, the solutions are x\equiv 5, 8, 20\bmod 33 .
    Last edited by melese; September 27th 2010 at 06:25 AM. Reason: One solution overlooked.
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