My Math Forum Factorization and GCD for Rijndael's finite field

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 August 27th, 2017, 07:27 AM #1 Newbie   Joined: Jun 2009 Posts: 19 Thanks: 1 Factorization and GCD for Rijndael's finite field I'm factoring polynomials in GF(2^8) with modulo polynomial $\displaystyle m = 2^8 + 2^4 + 2^3 + 2^1 + 2^0$ In particular, I factored a = 0x49 = $\displaystyle 2^6 + 2^3 + 2^0 = 2^1 * (2^1 + 2^0)^6 * (2^2 + 2^1 + 2^0) * (2^4 + 2^1 + 2^0) * (2^3 + 2^1 + 2^0)$ (mod m) b = 0x64 = $\displaystyle (2^6 + 2^5 + 2^2) = (2^1)^3 * (2^1 + 2^0) * (2^3 + 2^1 + 2^0) * (2^3 + 2^2 + 2^0)$ (mod m) these numbers are multiplicative inverses and I can calculate directly using long division that ab(mod m) = 2^0. I assume GCD should also be 1, i.e. 2^0. Now I want to calculate GCD using just factored irreducible polynomials. I know that for integers GCD equals the product of prime numbers (including their powers) present in both factorization, does it still hold here? $\displaystyle gcd = 2^1 * (2^1 + 2^0) * (2^3 + 2^1 + 2^0)$ (mod m) = $\displaystyle 2^5 + 2^4 + 2^3 + 2^1$ Can anyone please tell me where I am mistaken? Thank you!

 Tags factorization, field, finite, gcd, rijndael

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