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Math Help - find solution set for this linear congruence

  1. #1
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    find solution set for this linear congruence

    x^2-2 is congruent to 0(mod7)
    Last edited by jmedsy; November 2nd 2009 at 07:58 PM.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by jmedsy View Post
    x^2-2 is congruent to 0(mod7)
    Note that 3^2-2=7=0 (mod\ 7) and 4^2-2=14=0 (mod\ 7).
    So x=3 (mod\ 7) or x=4 (mod\ 7).
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    thanks
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    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by jmedsy View Post
    That's true, but its also true for x = 10. How can I solve for all possible x?
    10 falls into the solution set x=3 (mod\ 7).
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    Quote Originally Posted by alexmahone View Post
    10 falls into the solution set x=3 (mod\ 7).
    Once you had determined that x=3 or x=4 satisfied the congruency, did you continue to check congruency for greater x? How did you know when to stop?
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by jmedsy View Post
    Once you had determined that x=3 or x=4 satisfied the congruency, did you continue to check congruency for greater x? How did you know when to stop?
    You need to check congruency only for x ranging from 0 to 6. (If the modulus was n, you should have checked for x ranging from 0 to n-1.)
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  7. #7
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    right, that makes sense. Is there any correlation between the degree of that polynomial and the number of different modular congruences between 0 and 6?
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