If $a\in GF(2^n)$, then there is the element in $GF(2^n)$ such that $x^2 = a$. Why?
2026-04-07 12:57:19.1775566639
Why are the elements of GF(q), whose characteristic is 2, all squares?
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For a finite field $F$, the multipülicative group $F^\times$ is cyclic of order $|F|-1$.