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Thread: help with modular arithmetic please

  1. #1
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    help with modular arithmetic please

    Hi all,

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


    It is known that

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

    from these we can say that if

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

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

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

    From reasonably extensive computer modelling of n, it also appears that $\displaystyle n\equiv0\ mod\ 3$ 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
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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.

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

    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:

    $\displaystyle n\equiv\ 0,3,6,9\ mod\ 12$
    which then proves that:
    $\displaystyle n\equiv\ 0\ mod\ 3$
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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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