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I am looking for an elementary evaluation (if one exists) of the exponential sum

$$ G_r(a,b) = \sum_{x \in \mathbb{F}_{2^r}} \psi(ax^2 + bx), $$

where $a,b \in \mathbb{F}_{2^r}^*$ are both units, $\psi(x) = e(Tr(x)/2)$ and $Tr : \mathbb{F}_{2^r} \to \mathbb{F}_2$ is the usual field Trace map

$$ Tr(x) = \sum_{i=0}^{r-1} x^{2^i}. $$

It should be noted that

$$ G_r(a,0) = G_r(0,a) = 0, $$

since the map $x \mapsto x^2$ permutes the elements of $\mathbb{F}_{2^r}$.

I feel that such a sum must have surely been studied before, but I am having trouble both evaluating the sum and finding references for it. Short of an explicit formula for the sum, any information (or any reference to where this sum might be studied) would be appreciated. I found no information on this sum in the usual suspects: Ireland-Rosen, Iwaniec-Kowalski and "Gauss and Jacobi Sums," by Berndt, Evans, and Williams.

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It should also be noted that the method for evaluating such sums over finite fields of odd characteristic is to complete the square, which is not applicable in the above case. –  David Sep 14 '11 at 3:49
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. –  Noam D. Elkies Sep 14 '11 at 3:57
@Noam. Thank you for your insight. If you add your comment as an answer, I would love to accept it. –  David Sep 14 '11 at 4:20
@David: you're welcome & thanks & that was quick :-) –  Noam D. Elkies Sep 14 '11 at 4:37
The technique generalizes to linearized polynomials, i.e. polynomials where only terms of degrees that are power of the characteristic occur. This technique is relatively common in finite fields, and coding theorists often use this trick, when studying quadratic forms in char 2. The classic tome Finite Fields by Lidl & Niederreiter (Cambridge Univ. Press) is the reference. –  Jyrki Lahtonen Sep 14 '11 at 7:09
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up vote 18 down vote accepted


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 above: 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.

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