Fibre sizes for functions on finite fields

$\newcommand{\fQ}{\mathbb F_Q}\newcommand{\fQs}{\fQ\setminus\{0\}}$Claim 1 is answered at this related question, so we care about Claim 2. In addition to the notation from the question, we set $q=p^k$, $Q=p^n$, $n_i=|Y_i|$. As noted there, we have \begin{equation} Q-1=n_1+2n_2+(p^d+1)n_3. \end{equation} With $f:\fQs\mapsto\fQ$, $x\mapsto x^q-1/x$, let $v_r$ be the number of subsets $\{x_1,\dots,x_r\}\subseteq\fQs$ of size $r$ with $f(x_i)=f(x_j)$ for all $i,j$. Then clearly \begin{align*} v_2 &= 2n_2+p^d(p^d+1)n_3\\ v_3 &= (p^d-1)p^d(p^d+1)n_3. \end{align*} Note that once we know the values of $v_2$ and $v_3$, we get $n_1,n_2,n_3$ from these three linear equations in triangular form.

Lemma 1. For $x\in\fQs$ and $z\in\fQ\setminus\{0,1\}$, the following are equivalent:

1. $f(x)=f((1-z)x)$.
2. $z^q-z^{q-1}=1/x^{q+1}$.

Proof. A straightforward computation, using $(1-z)^{q+1}=(1-z)^q(1-z)=(1-z^q)(1-z)$.

From now on we add the assumptions from Claim 2, namely that $p=2$ and that $n/d$ is odd.

Lemma 2. $v_2=2^n-2$.

Proof. We need to count the number of pairs $(x,y)$ where $x,y\in\fQs$, $x\ne y$, and $f(x)=f(y)$. Write $y=(1-z)x$, where $z\in\fQ\setminus\{0,1\}$. By Lemma 1, this is equivalent to $z^q-z^{q-1}=1/x^{q+1}$. As $q+1$ is relatively prime to $Q-1$ (see the lemma at related question), the map $X\mapsto X^{q+1}$ is bijective on $\fQ$. Therefore we get the same count if we replace $1/x^{q+1}$ with $x$. Thus, as any $z\in\fQ\setminus\{0,1\}$ gives a nonzero $x$, we obtain $v_2=Q-2$.

Lemma 3. $v_3=2^n-2^d$.

Proof. Now let $x$, $(1-z_1)x$, and $(1-z_2)x$ be three distinct nonzero elements from $\fQ$ which are mapped to the same element under $X\mapsto X^q-1/X$. Again, by Lemma 1, this is equivalent to $z_i^q-z_i^{q-1}=1/x^{q+1}$ for $i=1,2$.

As $q+1$ is relatively prime to $Q-1$ (see above), we may replace $1/x^{q+1}$ by $x$. For any $z_1\not\in\{0,1\}$ we get a unique nonzero $x$. Thus, we need to count the pairs $(z_1,z_2)$ where $z_1,z_2\not\in\{0,1\}$, $z_1\ne z_2$, and $z_1^q-z_1^{q-1}=z_2^q-z_2^{q-1}$. Write $z_2=\delta z_1$ (so $\delta\not\in\{0, 1\}$.) This yields (similarly as in the proof of Lemma 1) $z_1=\frac{\delta^{q-1}-1}{(\delta-1)^q}$. This gives a nonzero $z_1$ if and only if $\delta^{q-1}\ne1$. As $\gcd(q-1,Q-1)=p^d-1$, the number of choices for $\delta$ is $Q-1-(p^d-1)=2^n-2^d$.

From the three linear equations (which are already in triangular form) we immediately get the expected numbers $|Y_i|$.

Remark. A slight modification allows to cover the case $p>2$ and $n/d$ odd. Then $\gcd(q+1,Q-1)=2$. So similarly as in the proof of Lemma 2 we now get $z^q-z^{q-1}=1/x^2$. As $q-1$ is even, $z^{q-1}$ is a square. So the condition on $z$ is that $z-1$ is a square. The condition $z\ne0,1$ gives $(Q-3)/2$ or $(Q-1)/2$ possibilities for $z$, depending on whether $-1$ is a square in $\fQ$ or not. For each admissible $z$ there are two possibilities for $x$. Thus $v_2=Q-3$ if $Q\equiv 1\pmod 4$, and $v_2=Q-1$ if $Q\equiv -1\pmod 4$.

Similarly, one can compute $v_3$. The calculation as in the proof of Lemma 3 yields $z_1-1=-\left(\frac{\delta}{\delta-1}\right)^{q-1}$. Recall that $z_1-1$ need to be a square. So $v_3=0$ if $Q\equiv -1\pmod 4$. If, however, $Q\equiv 1\pmod 4$, then this condition on $z_1$ is automatically fulfilled. As there are two possibilities for $x$ for each $z_1$, we obtain $v_3=2(p^n-p^d)$ in this case.