Is the absolute trace of $\frac{xy}{(x+y^4)^2}$ equal to zero?

74 Views Asked by At

Let $\mathbb{F}_{q}$ with $q=2^m$ be the finite field of $q$ elements. Let $x,y \in \mathbb{F}_{q}$. Let $m$ be any integer. If $xy(x+y)=1$ and $x+y^4\neq 0$, then experiment data for $m=3,4,5,6,7,8,9,10$ suggest that the absolute trace function from $\mathbb{F}_{q}$ to $\mathbb{F}_{2}$ of the following element: $$\frac{xy}{(x+y^4)^2}$$ is equal to zero.

I tried to write the above element as $z+z^2$ with $z\in \mathbb{F}_{q}$, but it seems that it is not easy.