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Math Help - help with modular arithmetic please

  1. #1
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    help with modular arithmetic please

    Hi all,

    n=((a^2+b^2)(c^2+d^2)(f^2+2ef-e^2))^2+((a^2+b^2)(d^2+2cd-d^2)(e^2+f^2))^2+(b^2+2ab-a^2)(c^2+d^2)(e^2+f^2))^2


    It is known that

    a^2+b^2\equiv 1\ mod\ 4
    c^2+d^2\equiv 1\ mod\ 4
    e^2+f^2\equiv 1\ mod\ 4

    from these we can say that if

    a\equiv(0\ mod\ 4)\ or\ (2\ mod\ 4)\ then\ b\equiv\ (1\ mod\ 4)\ or\ (3\ mod\ 4) or vice versa.

    The same is true for the relationship between (c and d) and (e and f)

    I have found that n\equiv\ 3\ mod\ 4 by using permutations of the above.

    From reasonably extensive computer modelling of n, it also appears that n\equiv0\ mod\ 3 but I can't seem to relate that to the above. Is it obvious why and I'm just suffering brain-freeze?

    Thanks for helping thaw me out
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  2. #2
    Junior Member
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    Re: help with modular arithmetic please

    Its pretty late, and I haven't really thought it all the way through but since 12 is the lcm of 3 and 4. Maybe you could convert the entire thing to mod 12.

    a^2+b^2\equiv\ 1,5,9\ mod\ 12
    c^2+d^2\equiv\ 1,5,9\ mod\ 12
    e^2+f^2\equiv\ 1,5,9\ mod\ 12

    Then build a relation table for a and b mod 12. Using a computer program you might be able to handle all the permutations, to find out what possible values n can hold mod 12. Of course you are expecting to see:

     n\equiv\ 0,3,6,9\ mod\ 12
    which then proves that:
    n\equiv\ 0\ mod\ 3
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  3. #3
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    Re: help with modular arithmetic please

    Thanks takatok, I can do that. I was just hoping for something more elegant.
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