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Thread: need help by tonight PLEASE!!!

  1. #1
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    need help by tonight PLEASE!!!

    How do I solve these kind of quadratic congruence problems?!

    x^2 is congruent to 53 mod 143

    I said 143=11x13

    so x^2 is congruent to 53 mod 11 AND x^2 is congruent to 53 mod 13


    where do i go from there?! Please any help would be great..My final is tomorrow and this is the only thing that is confusing me!!
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  2. #2
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    Quote Originally Posted by kel1487 View Post
    How do I solve these kind of quadratic congruence problems?!

    x^2 is congruent to 53 mod 143

    I said 143=11x13

    so x^2 is congruent to 53 mod 11 AND x^2 is congruent to 53 mod 13


    where do i go from there?! Please any help would be great..My final is tomorrow and this is the only thing that is confusing me!!
    That's the right way to start. Then 53\equiv 9\pmod{11}, and the equation x^2\equiv9\pmod{11} has solutions x\equiv\pm3\pmod{11}. Also, 53\equiv 1\pmod{13}, and the equation x^2\equiv1\pmod{13} has solutions x\equiv\pm1\pmod{13}.

    So we are looking for numbers of the form (multiple of 11) ±3, that are also of the form (multiple of 13) ±1. The easiest way to find such numbers is simply to go through the multiples of 11, looking at the numbers ±3 from these, and seeing if they are within ±1 of a multiple of 13.

    The first one we find is 14 (=11+3=13+1).

    Then 25 (=22+3=26-1); 118 (=121-3=117+1); and 129 (=132-3=130-1).

    So there are four solutions, x\equiv 14, 25, 118, 119\pmod{143}.
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