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2

To answer this, let's look at what the elements of $\mathbb{F}_{2^4}$ are: polynomials with coefficients in $\mathbb{F}_2$ that are reduced modulo $f(x)=x^4+x+1$. This reduction means that any power of $x$ greater than or equal to $x^4$ can be rewritten as a sum of smaller powers: in particular, $x^4=x+1$, and higher powers we can factor this $x^4$ out ...

4

Nothing wrong with the other answers. I just want to point out that if you are using a normal basis representation of the field in question, then there is a very efficient way of calculating the square root. You recall that squaring of the element is equivalent to cyclically rotating its coordinates w.r.t. the normal basis. Therefore to calculate the square ...

5

To complete poncho's answer, if you know some Galois theory. The map $\sigma: x\mapsto x^2$ from $\mathbf{F}_{2^n}$ to itself is simply the Frobenius automorphism (relative to $\mathbf{F}_2$). It generates the Galois group $\mathrm{Gal}\left(\mathbf{F}_{2^n}/\mathbf{F}_2\right)$, which is cyclic of order $n$, and so its inverse (which is, by definition, the ...

5

Yes, there is an algorithm for efficiently computing square roots in $GF(2^n)$. I don't know if this is the most efficient known, but the existence of an efficient algorithm can be shown by observing that squaring within $GF(2^n)$ is a bitwise linear operation, hence it is equivalent to taking the bit representation of the value, and multiplying it by an ...

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