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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.

Sunshning
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    Over $\Bbb{F}8$ all the solutions of $xy(x+y)=1$ are gotten by applying the symmetries (permute $x,y,x+y$ any which way you want) to a single solution. Over $\Bbb{F}{16}$ the rational points of this elliptic curve all have their coordinates in $\Bbb{F}_4$ (see the last example in this old answer of mine, it's really the same curve after the obvious transformation via homogeneous coordinates). This made me suspect that we might be seeing something explained by the law of small numbers. – Jyrki Lahtonen Feb 16 '23 at 07:21
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    (cont'd) Often facts like this can either be proved relatively simply, or counterexamples can be found in large enough fields via some use of exponential sums. I don't see a way of turning this into an exponential sum where we could use Weil-type bounds, and the OP has checked this for bigger fields anyhow. – Jyrki Lahtonen Feb 16 '23 at 07:28
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    (cont'd) Another curious fact is the lack of symmetry in the rational function. It is possible to turn the problem into a more symmetric statement, making the equivalent claim that the quartic we get by multiplying the two quadratic equation related to this trace condition and its $x\leftrightarrow y$ counterpart, when we can claim that the resulting quartic splits completely over $\Bbb{F}_q$. The added symmetry allows us to reduce the number of variables (replacing the equation with a trace condition), but I don't see a way of proving that version either. – Jyrki Lahtonen Feb 16 '23 at 07:31

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