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I am curious how I would go about solving this problem. I would show some work but I have no idea where I would start with this problem. If someone could give me any direction it'd be greatly appreciated, or even just a source I can read that will point me in the right direction. I've been trying to do some research on my own but many of the sources seem to be different than what is happening here.

Thank you so much

user3371137
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    Just compute the remainder of the polynomials mod $f$. – lhf Oct 10 '19 at 21:47
  • @lhf so in this case if I were to compute A^2 for example I would first compute A mod f and then after that I would square A mod f to get the result A^2? – user3371137 Oct 10 '19 at 21:49
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    Since $\deg(A) \lt 8$, you already know $A \pmod{f}$. You also know (via long division) that $A^2 = q(x)f(x)+r(x)$ for some $r(x)$ with $\deg(r) \lt 8$, so $A^2 \equiv r \pmod{f}$. – Robert Shore Oct 10 '19 at 21:56
  • Sorry i've been trying to decipher your instructions for awhile now I'm still not following though could you maybe explain it in a different way? – user3371137 Oct 11 '19 at 00:10
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    Multiply $A$ times itself, remembering that the coefficients are in $\Bbb Z/2\Bbb Z$. Then take your product, and apply Euclidean Division to it to get the remainder, which is your answer. – Lubin Oct 11 '19 at 01:51
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    What Robert Shore and Lubin said. I want it on record that the formulation of the exercise is not optimal. The elements of $GF(2^8)$ are not polynomials. Rather they are cosets of polynomials modulo $f(x)$. Or (my preferred way) polynomials evaluated at a zero $\alpha$ of $f(x)$. Also, the verb Solve is misleading. There is no equation to solve here. Just calculations. This is what we get, when somebody tries to teach innocent students finite fields without going via quotient rings :-( – Jyrki Lahtonen Oct 11 '19 at 04:20
  • Thank you very much guys this helped a lot it took me a while but I managed to solve it I appreciate all the guidance – user3371137 Oct 11 '19 at 11:16

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