ag.algebraic geometry - Existence of hyperelliptic curve with specific number of points in a family

I think that this is equivalent to a known open question. Here are
the details. For $K:=mathbb{F}_{2^n}$, the function $f:ymapsto
y+y^2:Kto K$ is $mathbb{F}_2$-linear, and its kernel ${0,1}$ has
dimension 1. The image is therefore of dimension $n-1$, and for $z$ in
the image, the fiber $f^{-1}(z)$ has exactly 2 elements.

Hence, to prove that $y^2+y=x^k+ax$ has exactly $2^n$ solutions
for some fixed $ain K$, we have to show that $|{xin K|x^k+axin
mathrm{Im}(f)}|=2^{n-1}$.

Since $sigma:ymapsto y^2$ is a generator of the Galois group of
$K/mathbb{F}_2$, Hilbert's Theorem 90 (in additive form) says that
$zin mathrm{Im}(f)$ if and only if $mathrm{Tr}(z)=0$, where
$mathrm{Tr}$ stands for the trace map from $K$ to $mathbb{F}_2$.

So the problem is equivalent to showing that there exists an $aneq 0$ in $K$
such that $|{xin K|mathrm{Tr}(x^k+ax)=0}|=2^{n-1}$. In other
words, we would like to show that there exists a nonzero $ain K$ such that
$$
S_k(a):=sum_{xin K}(-1)^{mathrm{Tr}(x^k+ax)}=0.
$$

Apparently, this question was addressed in the coding community. In
detail, in [1, p. 258], the following conjecture (of Helleseth) is
mentioned:

``Conjecture 3. For any $m$ and $k$ such that
$mathrm{gcd}(2^m-1,k)=1$, the sum
$sum_{xinmathbb{F}_{2^m}}(-1)^{mathrm{Tr}(x^k+ax)}$ is null for at
least one nonzero $a$.'' (Note that $n$ in the current question is
$m$ in 1).

It seems that in [1, Corollary 1, p. 253], Conjecture 3 is proved for
even $m$ and for certain values of $k$ (the ``Niho exponents,''
defined on p. 252 of 1).

Interestingly, at least at a first glance it seems that 1 has
nothing to say on $kin{1,ldots,2^{n-1}}$, but to me it seems that
this case is trivial (am I missing something?): Consider a normal
basis for $K/mathbb{F}_2$, that is, a basis $B$ consisting of an orbit of
an element $gammain K$ under the Galois group of $K/mathbb{F}_2$
(the $i$th element of $B$ is $b_i:=gamma^{2^i}$ for
$iin{0,ldots,n-1}$).

From the linearity of the trace and the fact that the trace is onto,
we must have $mathrm{Tr}(b)=1$ for at least one element $bin B$, and
from $mathrm{Tr}(b^2)=mathrm{Tr}(b)$ we then have
$mathrm{Tr}(b)=1$ for all $bin B$. So the trace of an element in $K$
is just the modulo-2 sum of the coefficients in its decomposition
according to the basis $B$.

Let $a$ be any element in the
trace-dual basis of $B$, say $mathrm{Tr}(ab_i)=delta_{i,0}$. Then
for $k=2^j$, if we write $x=sum_i alpha_i b_i$, we
get: $mathrm{Tr}(ax)=alpha_0$,
$mathrm{Tr}(x^k)=mathrm{Tr}(x)=sum alpha_i$ (sum in
$mathbb{F}_2$). These agree for half of the $xin K$, as required.

That's about it. I hope at least some of this makes sense :) I also
hope that the original person asking this question didn't actually want
to solve the above open question by converting it to a question about
curves, for then this answer is useless.

1 P. Charpin, ``Cyclic codes with few weights and Niho exponents,''
Journal of Combinatorial Theory, Series A 108 (2004)
247--259.