how $𝑎^𝑒𝑢^𝑒 +𝑎^{𝑒−1}𝑢^{𝑒−1}𝑏+𝑎𝑢𝑏^{𝑒−1} +𝑏^𝑒$ is derived from $(𝑎𝑢+𝑏)^𝑒$ in $F_{2^{2𝑛}}$?

We represent an element $$𝑥$$ of $$F_{2^{2𝑛}}$$ as a linear polynomial $$𝑥 = 𝑎𝑢 + 𝑏$$ over $$F_{2^𝑛}$$ with multiplication modulo the irreducible polynomial $$𝑢^2 + 𝑢 + 1$$.

Note that $$e=2^{2𝑘}+1$$ and $$𝑢^2 = 𝑢+1,𝑢^4 = 𝑢,...,𝑢^{2^2𝑘} = 𝑢,𝑢^{2^{2𝑘}+1} = 𝑢+1$$. Then, by linearity of $$𝑥 ↦ 𝑥^{𝑒−1}$$

$$𝑥^𝑒 =(𝑎𝑢+𝑏)^𝑒 =𝑎^𝑒𝑢^𝑒 +𝑎^{𝑒−1}𝑢^{𝑒−1}𝑏+𝑎𝑢𝑏^{𝑒−1} +𝑏^𝑒$$

$$= (𝑎^𝑒 +𝑎^{𝑒−1}𝑏+𝑎𝑏^{𝑒−1})𝑢+𝑎^𝑒 +𝑏^𝑒$$

$$= (𝑏^𝑒 +(𝑎+𝑏)^𝑒)𝑢+𝑎^𝑒 +𝑏^𝑒$$

My question is how $$𝑎^𝑒𝑢^𝑒 +𝑎^{𝑒−1}𝑢^{𝑒−1}𝑏+𝑎𝑢𝑏^{𝑒−1} +𝑏^𝑒$$ is derived from $$(𝑎𝑢+𝑏)^𝑒$$ in $$F_{2^{2𝑛}}$$ by pen and paper?

• What is $e$? Something arbitrary? – Morgan Rodgers Apr 4 at 15:08
• e=$2^{2𝑘}+ 1$ . – hardyrama Apr 4 at 15:34
• Then since $x \mapsto x^{e-1}$ is linear, $(au+b)^e = (au+b)^{e-1}(au+b) = (a^{e-1}u^{e-1}+b^{e-1})(au+b)$. – Morgan Rodgers Apr 4 at 16:19
• thanx for the answer – hardyrama Apr 4 at 16:30
• Add the definition of $e$ to the question body, please. Otherwise the question is unclear. – Jyrki Lahtonen Apr 4 at 21:34