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Math Help - incongruent solutions to (mod n)

  1. #1
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    incongruent solutions to (mod n)

    7x = = 3 (mod 15)
    take the gcd (7,15) = 1 therefore 1 and 1/3 are 2 incongruent solutions??

    6x = = 5 (mod 15)
    gcd = 3 but 3 isn't a divisor of 5 therefore there are no solutions??

    x^2 = = 1 (mod8) I get 4 and 1 as two incongruent solutions??

    I just want to see if my working is accurate...

    THANK YOU
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by duggaboy View Post
    7x = = 3 (mod 15)
    take the gcd (7,15) = 1 therefore 1 and 1/3 are 2 incongruent solutions??
    Since gcd(7,15)=1, you can find y such that 7y=1 mod 15. After that, multiply the whole by 3 in order to get x.

    Euclidian algorithm, to find y :

    15=7x2+1 --> -7x2=1-15=1 mod 15 (remember, i told you you could add as many times as you want 15).

    -2 mod 15=-2+15 mod 15=13 mod 15

    Therefore, y=13

    Multiplying by 3 : x=39=9 mod 15

    6x = = 5 (mod 15)
    gcd = 3 but 3 isn't a divisor of 5 therefore there are no solutions??


    x^2 = = 1 (mod8) I get 4 and 1 as two incongruent solutions??
    Hmmm what is "incongruent" ?

    1=1 mod 8

    whereas 4=0 mod 8
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  3. #3
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    -2 mod 15=-2+15 mod 15=13 mod 15

    I don't know where this line came from?

    x^2 = = 1 (mod8) I get 4 and 1 as two incongruent solutions??

    Should I be setting x = to an actual integer or just giving a generalization?
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  4. #4
    Moo
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    Quote Originally Posted by duggaboy View Post
    -2 mod 15=-2+15 mod 15=13 mod 15

    I don't know where this line came from?
    Remember ? When dealing with mods, you can add as many times 15 as you want.
    So you can add 15, it'll just be the same.
    By convention, we often write a positive number, inferior to n (here, 15) in the congruence. That's what I did for -2.

    x^2 = = 1 (mod8) I get 4 and 1 as two incongruent solutions??

    Should I be setting x = to an actual integer or just giving a generalization?
    You know that every number is either =0,1,2,3,4,5,6 \ or \ 7 \mod 8, so you can try them out.
    But 1 verify the equation, while 2 and 6 don't

    When you find that for example 2 is not a solution, you can say that "any number in the form 2+8k, k in Z, is not solution of the equation"
    Last edited by Moo; June 3rd 2008 at 10:48 AM.
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  5. #5
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    Thank You So Much
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