TITLE: A quadratic form
QUESTION [7 upvotes]: Let $q$ be a power of 2.  Let $P$ be the set of  polynomials in 
$F_q [x]$ of degree d or less.
Let $\mathbb{Z}$ be the ring of integers. For any $f \in P$, let $\psi(f)$ be the
number of distinct roots of $f$ in $F_q$. Note that $\psi(0) = q$.
For any map  $A$ from $P$ to $\mathbb{Z}$, one can  compute the summation
  $$\sum_{f, g \in P} A(f)A(g) \psi (f+g).$$
My question is:
what is the minimum positive value of the summation?
For $d=0,1$, the minimum is $q$. What happens if $d$ is bigger?
I am especially interested in the case when d=q/2-1.
Thanks a lot,
Qi

REPLY [2 votes]: (Inspired from Charles Siegel's answer) what you can do for the $d=\frac{q}{2}$ case is:
Divide the $q$ elements of $F_q/\{0\}$ into two parts $\{\alpha_{1},\dots, \alpha_{\frac{q}{2}}\}$ and $\{\beta_1,\dots,\beta_{\frac{q}{2}-1}\}$ so that $\prod \alpha_i\neq \prod \beta_j$. Denote $S(x)=\prod (x-\alpha_i)$ and $T(x)=\prod(x-\beta_j)$. Let $r_1(x)+r_2(x)=S(x)+T(x)$ be arbitrary and $f_1=S+r_1,f_2=S+r_2$. Now we let $A(f_1)=A(f_2)=1,A(r_1)=A(r_2)=-1$ and all other $A(g)=0$. The quadratic form has the value
$$-2\psi(f_1+r_1)-2\psi(f_1+r_2)-2\psi(f_2+r_1)-2\psi(f_2+r_2)+4\psi(r_1+r_2)+4q=4$$
I don't know if this can be lowered to 2...