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Math Help - Calculation in Rijndael field GF(2^8) and (mod x^4+1)?

  1. #1
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    Calculation in Rijndael field GF(2^8) and (mod x^4+1)?

    I have a 6th degree polynomial with unknown coefficients, but I need to take it (mod x^4+1) to find some 3rd degree polynomial. Using wolfram alpha and mathematica 9.0, I could only reduce the polynomial to a different 6th degree polynomial that was shorter.


    e.g. a_1x^6 + a_2x^5 + ... a_6x + a_7 is what I'm given, where each a_i is something like:
    a_1 = s_1 + s_3 + s_4
    a_2 = s_1 + s_2 + s_3
    a_3 = s_3 + s_4
    ...

    for some unknown s_i.

    Also, the coefficients are to (mod 2).

    How can I find a_1x^6 + a_2x^5 + ... a_6x + a_7 (mod x^4+1) as a 3rd degree polynomial?

    I heard that under mod (x^4 + 1) and (mod 2), I have congruence x^4 = -1 = 1 but that doesn't make sense to me?
    Last edited by HowDoIMath; November 18th 2013 at 10:48 AM.
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  2. #2
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    Re: Calculation in Rijndael field GF(2^8) and (mod x^4+1)?

    Since x^4+1 = 0 mod (x^4+1), then x^4 = -1 mod (x^4+1).

    Let me rewrite your 6th degree polynomial as:

    f(x) = a6 * x^6 + a5 * x^5 + ...a1 * x + a0 = (a6,a5,a4,a3,a2,a1,a0) ...in other words concentrate on the coefficients.

    Also note the subscript on the coefficient matches the power of x.

    Now a4 * x^4 = a4 * (-1) mod (x^4+1). Likewise a5*x^5 = a5*(-x) and a6*x^6 = a6*(-x^2).

    So (a6,a5,a4,a3,a2,a1,a0) = (a3,a2,a1,a0) - (a6,a5,a4) = (a3,a2-a6,a1-a5,a0-a4).

    So the cubic is: a3 * x^3 + (a2-a6)*x^2 + (a1-a5)*x + (a0-a4).

    If you want to work in GF(2^8), just take the coefficients mod (2^8).
    Thanks from topsquark and HowDoIMath
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  3. #3
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    Re: Calculation in Rijndael field GF(2^8) and (mod x^4+1)?

    Thanks a lot qmech! It's pretty obvious once you see how it's done...
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