math.NT — Number Theory
The Welch function is not third-order sum-free
Let be odd and let $\F_{2^n}$ be the field with elements. A function on $\F_{2^n}$ is third-order sum-free when its values over any eight elements forming a three-dimensional affine subspace never add up to zero. The Welch function is the power map ; it is one of the classical almost perfect nonlinear functions, and Hou and Zhao conjectured that for it is not third-order sum-free. We prove this. The question comes down to a linearised polynomial in whose roots form a vector space over $\F_2$ containing and , and the Welch function fails to be third-order sum-free exactly when has dimension three for some outside $\F_2$. We show that the dimension is always or , and that it is if and only if two explicit trace conditions hold: with , both $\Tr(1/(x+u))$ and $\Tr(xu/(x+u))$ equal . The proof applies the field automorphism , whose square is the Frobenius map, to turn into an additive polynomial of degree eight whose roots always lie in $\F_{2^n}$. Counting the elements that satisfy both trace conditions gives exactly of them, where is the binary Kloosterman sum of $\F_{2^n}$, and Weil's bound makes this count positive for every . The Welch function sums to zero over exactly three-dimensional linear subspaces, and the count agrees with the numbers that Hou and Zhao computed for . Every statement is proved by hand.