# How to caculate the inverse of function $x^3$ in $\mathbb{F}_{2^n}$

How to caculate the inverse of function $$x^3$$ in $$\mathbb{F}_{2^n}$$?, Any monomial $$x^d$$ is a permutation in the field $$\mathbb{F}_{2^n}$$ iff $$gdc(d,2^{n}-1)=1$$,why?

• Welcome to Cryptography. $x^3$ is not a function but rather a polynomial representation of the elements of the field $\mathbb F_{2^2}$. The default way to find the inverse is using extended-gcd on the polynomials. If you only need the result, use SageMath. Is this HW question? May 23, 2022 at 10:31
• @kelalaka: actually, I believe that by $x^3$, he is talking about the function $F(z) = z \cdot z \cdot z$, which is well defined on any field, such as $\mathbb{Z}_{2^{n}}$. May 23, 2022 at 11:33
• @poncho your interpretation is better than mine. May 23, 2022 at 11:45

The order of the multiplicative group of $$\mathbb F_{2^n}$$ is $$2^n-1$$. If 3 is coprime to $$2^n-1$$ then there exists $$d\in [1,\ldots,2^n=1]$$ such that $$3d\equiv 1\pmod{2^n-1}$$. We can find such a $$d$$ using the extended Euclidean algorithm.
The function on $$\mathbb F_{2^n}$$ $$y\mapsto y^d$$ is then the inverse of the map $$x\mapsto x^3$$ since for $$x\in\mathbb F_{2^n}^\times$$ we have $$(x^3)^d=x^{3d}=x^1$$ (the case $$x=0$$ is obvious).
This means that $$x\mapsto x^3$$ is bijective and hence a permutation.
In the case where $$3|2^n-1$$, if $$x^3=y$$ in $$\mathbb F_{2^n}$$ then so to does $$(\omega x)^3=y$$ and $$(\omega^2 x)^3=y$$ where $$\omega$$ is a cube root of 1 in $$\mathbb F_{2^n}$$. It follows that in this case the map is not injective and so not a permutation.