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Math Help - mod121

  1. #1
    Super Member dhiab's Avatar
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    mod121

    Solve in Z :  x^{2}+3x+5\equiv 0 (mod121)
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  2. #2
    Super Member Bacterius's Avatar
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    Hello, Dhiab,

    Solution : x^2 + 3x + 5 never divides 121, so this equation has no solutions in \mathbb{Z}

    Proof : x^2 + 3x + 5 \equiv (x \pmod{121})^2 + 3(x \pmod{121}) + 5 \pmod{121}. Apply an exhaustive search on x \in [0,120] (since values wrap around beyond this interval), and prove that none of the values of the quadratic modulo 121 are divisible by 121.

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  3. #3
    MHF Contributor

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    121= 11^2. Obviously (121)^2+ 3(121)+ 5 is congruent to 5 (mod 121) and any other number that divides 121 must also divide 11. So it is sufficient to do an "exhaustive search" of numbers in [0, 10].
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