Are there some solutions in $\mathbb{F}_{2^k}\times \mathbb{F}_{2^k}$ to the equation in $x,y$: $x^{2^t-1}=y+y^{2^t}$? Let $k$ be an odd positive integer and $t$ a positive integer with $t<k$ and $t|k$. Denote $\mathbb{F}_{2^k}$ the finite field with $2^k$ elements. I have a quenstion about the solutions in $\mathbb{F}_{2^k}\times \mathbb{F}_{2^k}$ to the following equation in $x,y$:$$x^{2^t-1}=y+y^{2^t}$$
According to some expriements with Magma for $k=9,15,21$, the above equation in $x, y$ has some solutions $(x,y)$ with $x\neq 0$ in $\mathbb{F}_{2^k}\times \mathbb{F}_{2^k}$. Thus, I guess that the equation always has some solution $(x,y)$ in $\mathbb{F}_{2^k}\times \mathbb{F}_{2^k}$ with $x\neq 0$ . Can you help me to give an answer to this quenstion?
 A: A case not covered by Jyrki Lahtonen

If some prime $p$ divides $k$ but not $t$ then yes there is a solution for $x^{2^t-1}=y+y^{2^t}$ in $\Bbb{F}_{2^k}^*\times \Bbb{F}_{2^k}^*$.

Take any $y\in \Bbb{F}_{2^p}-\Bbb{F}_2$.
$y^{2^t}+y\in \Bbb{F}_{2^p}^*$ so $(y^{2^t}+y)^{2^p-1}=1$.
$\gcd(2^p-1,2^t-1)=2^{\gcd(p,t)}-1=1$ and $2^p-1| 2^k-1$ so $2^p-1| \frac{2^k-1}{2^t-1}$ and hence $$(y^{2^t}+y)^{(2^k-1)/(2^t-1)}=1 $$ from which $\exists x\in \Bbb{F}_{2^k}^*, x^{2^t-1}=y^{2^t}+y$.
A: The claim may be true more generally, but at this time I can only prove it under the extra assumption that $\gcd(2^t-1,k/t)=1$. Under that assumption I can also derive a formula for the number of solutions such that $x\neq0$.

Let $\ell=k/t$, $K=\Bbb{F}_{2^k}$ the bigger field, and $F=\Bbb{F}_{2^t}$ the subfield.
Let's first look at the right hand side. The mapping $\sigma: K\to K, y\mapsto y+y^{2^t}$, is clearly $F$-linear. We have $\mathrm{Ker}(\sigma)=F$, so the image, call it $V$, is an $(\ell-1)$-dimensional subspace of $K$ over $F$. My general strategy is to show that each $1$-dimensional $F$-subspace of $K$ contains exactly one non-zero point of the form $x^{2^t-1}, x\in K$. As this applies to 1-dimensional subspaces of $V$ as well, we get the conclusion as well as a formula for the number of solutions.
Let us then look at the left hand side. The mapping $\delta:x\mapsto x^{2^t-1}$ is a $(2^t-1)$-to-$1$ mapping that sends all of $F^*$ to $1$. To get a handle on its image we need a bit of numerology. Let
$$
M=\frac{2^k-1}{2^t-1}=1+2^t+2^{2t}+\cdots+2^{(\ell-1)t}.
$$
We see that each term on the right has remainder $1$ modulo $2^t-1$, so
$$
M\equiv \ell\pmod{2^t-1}.
$$
It follows that if $\gcd(\ell,2^t-1)=1$ then also $\gcd(M,2^t-1)=1$. This is the key to my argument, for it implies that we can split $F^*$ off from $K^*$ as a direct sum component. By cyclicity of $K^*$ and the assumption on gcd, we have
$$
K^*=F^*\times S,\qquad(*)
$$
where $S$ is the (unique) cyclic subgroup of order $M$. So an arbitrary element $x\in K^*$ can be written as $x=zs$ with $z\in F^*$, $s\in S$, when
$$
x^{2^t-1}=z^{2^t-1}s^{2^t-1}=s^{2^t-1}\in S.
$$
Furthermore, as $\gcd(2^t-1,M)=1$, the restriction of $\delta$ to $S$ simply permutes its elements. Therefore $\mathrm{Im}(\delta)=S$.

The decomposition $(*)$ should be seen as saying that the subgroup $S$ parametrizes the $1$-dimensional $F$-subspaces of $K$. The properties of cyclic groups alone would imply that $x^{2^t-1}$ ranges over the subgroup $S$ even without the gcd-assumption. But the gcd-assumption is needed to make the intersection $S\cap F^*$ trivial.

That direct sum decomposition gives us the claim immediately.  If $\alpha\in K^*$ is arbitrary, then $\alpha =zs$ for uniquely determined $z\in F^*$, $s\in S$. It follows that $z^{-1}\alpha$ is the only element of the 1-dimensional subspace $F\alpha$ belonging to $S$

Still working under the assumption $\gcd(\ell,2^t-1)=1$ we see that $V$ intersects with $S$ in a set of size $(2^{t(\ell-1)}-1)/(2^t-1)$, one point of intersection for each 1-dimensional $F$-subspace of $V$. Taking into account the kernels of both $\sigma$ and $\delta$ we arrive at the formula
$$
N=2^t\cdot(2^t-1)\cdot\frac{2^{t(\ell-1)}-1}{2^t-1}=2^k-2^t
$$
for the number of solutions such that $x\neq0$. Of course, this could be checked with a computer.
A: Let $k=td$. Then $d\geq 3$ since $k$ is odd and $k>t$. For the case of $d=3$, the assertion holds by the arguments of Jyrki Lahtonen. Now, we will show that the desired assertion also holds true for the case of $d>3$.
Denote $\overline{\mathbb{F}_2}$ the algebraic closure of $\mathbb{F}_2$. Let PG$(3,\overline{\mathbb{F}_2})$ be the three-dimensional projective space.  Then the following proposition can be found in [1] or [2].
Proposition 1 Let $W$ be a surface and $H$ a plane of PG$(3,\overline{\mathbb{F}_2})$ such that $W\bigcap H$ contains a non-repeated
absolutely irreducible component defined over $\mathbb{F}_{2^k}$. Then $W$ possesses a non-repeated absolutely irreducible component defined over $\mathbb{F}_{2^k}$.
The following proposition can be found in [3] or [4].
Proposition 2 Let $L(x, y)$ be a polynomial in $\mathbb{F}_{2^k}[x, y]$ of degree $d$ and let $\#V\mathbb{F}_{2^k}^2(L)$ be the number of zeros of $L$ in $\mathbb{F}_{2^k}$. If $L$ has an absolutely irreducible component over $\mathbb{F}_{2^k}$, then

*

*$|\#V\mathbb{F}_{2^k}^2(L)-2^k|\le2^{\frac{k}{2}}(d-1)(d-2)+\frac{1}{2}d(d-1)^2+1,$
where the solutions at infinity is at most $d+1$.

Let $L(x,y): x^{2^t-1}=y+y^{2^t}$.
Our strategy is to show that $L(x,y)$ has an absolutely irreducible component over $\mathbb{F}_{2^k}$, then we use Proposition 2 to evaluate the numbers of the solutions to the equation: $L(x,y)=0$. Then we can show that under the assumption that $k>3t$, the numbers of the solutions with $x\neq 0$  are always greater than 0.
We can give the formula of $L$ in the three-dimensional projective space PG$(3,\overline{\mathbb{F}_2})$:
$$W\text{: }zx^{{2^i}-1}+yz^{{2^i}-1}+y^{2^i}=0.$$
The intersection of $W$ with the plane of equation $y=0$ is $zx^{{2^i}-1}=0$. The component $z=0$ is
non-repeated. Thus, $W$ possesses a non-repeated absolutely irreducible component defined over $\mathbb{F}_{2^k}$ according to Proposition 1. It means that $L$ has an absolutely irreducible
component over $\mathbb{F}_{2^k}$. For $x\not=0$, we can know that the number of solutions of $L$ without infinity at least
$$2^k-2^{t+1}-2^{\frac{k}{2}}(2^t-1)(2^t-2)-\frac{1}{2}2^t(2^t-1)^2-2.$$
by  Proposition 2 since $\#V\mathbb{F}_{2^k}^2(L(0,y))=2^t.$
The above number is equal to
$$(2^t-1)(2^t+2^{2t}+\cdots+2^{(\frac{k}{t}-1)t}-(2^{t+\frac{k}{2}}-2^{\frac{k}{2}+1})-(2^{2t-1}-2^{t-1})-1)-3.$$
Since $2^t-1\ge1$, we have that  if $k>3t$, then $2^{2t-1}-2^{t-1}<2^{2t}$ and $2^{t+\frac{k}{2}}-2^{\frac{k}{2}+1}<2^{(\frac{k}{t}-1)t}$. Hence, we conclude that if if $k>3t$, $\#V\mathbb{F}_{2^k}^2(L)-2^t>0$. This compltes the proof.
[1] Y. Aubry, G. McGuire, F. Rodier, A few more functions that are not APN infinitely
often, Finite Fields: Theory and Applications, 518 (2010) 23–31.
[2] D. Bartoli, T. Marco, On a conjecture on APN permutations, arXiv:2105.03702 (2021).
[3] X. Hou, Lectures on Finite Fields, vol. 190, American Mathematical Society, 2018.
[4] X. Hou, Applications of the Hasse–Weil bound to permutation polynomials, Finite Fields
and Their Applications, 54 (2018) 113–132.
