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Let $\eta =|A|/N$ and $g=1_A-\eta$. In the norm convention, the quadratic uniformity of a set meansIf uniformity is parametrized instead by $\Vert g{\Vert }_{U^3}^8\le \alpha$, replace the parameter below by ${\alpha }^{1/8}$. We prove the genuine-progression conclusion by placing nonnegative weights on a region where wrapping and repeated entries are impossible.

First suppose $N\ge 16$. Put $M=\lfloor N/16\rfloor$, $J=\{0,\dots ,M-1\}$, $t=M/N$, and let $w=1_J\ast 1_J$ be the normalized convolution on a finite group. Define the weighted countBoth weights are nonnegative. Their supports require $0\le x\le 2M-2$ and $M\le d\le 3M-2$ in the ordinary representatives. In particular $d\ge 1$ and $x+3d\le 11M-8<N$. Every contributing arithmetic progression is therefore a set of four distinct integers in $\{0,\dots ,N-1\}$, with no modular wrap.

The averages of both weights are $t^2$. The convolution theorem on a finite group and the Parseval identity on a finite group giveTranslation multiplies coefficients by phases, so $w(d-M)$ has the same Fourier coefficient sum. Telescoping the product of the four indicators around ${\eta }^4$ produces four error terms, each with one factor $g$ and all other factors bounded by one.

To bound a weighted error, expand both weights in their finite Fourier series. For frequencies $r,s$, the extra character is $e^{2\pi i(rx+sd)/N}$. Write it asand absorb these two characters into the functions at slopes zero and one. Even if the balanced factor is one of those functions, its $U^3$ norm remains at most $\alpha$ by the character invariance proved in part (i). The cyclic bound of part (ii), with $C_0=6^{1/8}$, therefore bounds each unweighted Fourier term by $C_0\alpha$. Summing absolute coefficients costs at most $t^2$. ConsequentlySince $t\ge 1/32$ and $\eta \ge \delta$, this is positive whenever $\alpha <{\delta }^4/(4096C_0)$. A positive count supplies the required genuine arithmetic progression. This is the argument that triangular weights count genuine four-term progressions; it avoids losing powers of $\log N$ by cutting off a sharp interval.

For completeness, treat $4\le N<16$. If $A$ is a proper nonempty subset, then $\eta (1-\eta )\ge \delta /N$. The norm bound from part (i) givesChoose $\alpha$ below this bound as well. Then quadratic uniformity forces $A={\mathbb{Z}}_N$, which contains $\{0,1,2,3\}$. Thus, for every $N\ge 4$, a sufficient threshold isThe restriction $N\ge 4$ is necessary: for $N=1,2,3$, the full group has balanced subset indicator zero and is quadratically uniform for every positive parameter, but no subset has four elements. The printed conclusion needs this elementary size qualification.