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Math Help - Diophantine equations

  1. #1
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    Diophantine equations

    Hi this is a hard thing. Solving Diophantine equations with many variables. My mind hang with this. Somebody helps me. This is it:


    x+y congruence to 1(mod7)
    x+z congruence to 2(mod7)
    y+z congruence to 3(mod7)
    Last edited by manusform; October 12th 2009 at 03:00 AM.
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  2. #2
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    Quote Originally Posted by manusform View Post
    Hi this is a hard thing. Solving Diophantine equations with many variables. My mind hang with this. Somebody helps me. This is it:


    x+y congruence to 1(mod7)
    x+z congruence to 2(mod7)
    y+z congruence to 3(mod7)

    Write x + y = 1 + 7k , x + z = 2 + 7m , y + z = 3 + 7n , k,m,n integers.
    Substract now the 3rd. one from the 1st.:

    x - z = -2 + 7(k-n) , and now add this to the second one

    2x = 7(k-n+m) ==> x = 0 (mod 7), and from here y = 1 (mod 7), z = 2(mod 7), and the general solution is:

    x = 7k
    y = 1 + 7m
    z = 2 + 7n

    k,m,n integers.

    Tonio
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  3. #3
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    x+y=1, x+z=2, y+z=3, so x+y+z=(1+2+3)/2=3. Since y+z=3, x=0. It follows that y=1 and z=2. (The same solution holds in modulo for any base.)
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