3 Typo correction aboce --> above

Trace is additive, and ${\rm Tr}(u)={\rm Tr}(u^2)$ for all $u$, so $ax^2+bx$ has the same trace as $(a+b^2)x^2$. Therefore the sum is $2^r$ if $a=b^2$ and zero otherwise.

In general, for a polynomial $P(x)$ over the field of $2^r$ elements, the sum of $\psi(P(x))$ is $2^r$ less than the number of affine points on the "hyperelliptic" curve $y^2+y=P(x)$. Here $P(x) = ax^2+bx$, so (for much the same reason I gave aboceabove: polynomials $\eta^2+\eta$ can be absorbed into $y^2+y$) the curve is rational, with $2^r$ points, unless $a=b^2$ when it is the union of two disjoint lines and has $2^{r+1}$ points.

2 Add geometrical picture

Trace is additive, and ${\rm Tr}(u)={\rm Tr}(u^2)$ for all $u$, so $ax^2+bx$ has the same trace as $(a+b^2)x^2$. Therefore the sum is $2^r$ if $a=b^2$ and zero otherwise.

In general, for a polynomial $P(x)$ over the field of $2^r$ elements, the sum of $\psi(P(x))$ is $2^r$ less than the number of affine points on the "hyperelliptic" curve $y^2+y=P(x)$. Here $P(x) = ax^2+bx$, so (for much the same reason I gave aboce: polynomials $\eta^2+\eta$ can be absorbed into $y^2+y$) the curve is rational, with $2^r$ points, unless $a=b^2$ when it is the union of two disjoint lines and has $2^{r+1}$ points.

1

Trace is additive, and ${\rm Tr}(u)={\rm Tr}(u^2)$ for all $u$, so $ax^2+bx$ has the same trace as $(a+b^2)x^2$. Therefore the sum is $2^r$ if $a=b^2$ and zero otherwise.