= Solution
Use $\Delta_hf(x)=f(x+h)\overline{f(x)}$ and the unnormalized <Fourier transform>
$$
\widehat g(r)=\sum_{x\in\mathbb Z_N}g(x)e(-rx/N).
$$
We state the eighth-power convention explicitly. A bounded function is $\alpha$-quadratically uniform if
$$
\boxed{\|f\|_{U^3}^8\le\alpha,\qquad
\|f\|_{U^3}^8=\mathbb E_{x,h_1,h_2,h_3}
\prod_{\epsilon\in\{0,1\}^3}\mathcal C^{|\epsilon|}f(x+\epsilon_1h_1+\epsilon_2h_2+\epsilon_3h_3),}
$$
where $\mathcal C$ means conjugation and every expectation is uniform. The expression is real and nonnegative. Applying <Parseval identity on a finite group> to the <multiplicative derivatives> gives the equivalent condition
$$
\|f\|_{U^3}^8=\frac1{N^5}\sum_{h,r}|\widehat{\Delta_h f}(r)|^4\le\alpha.
$$
This is the <Derivative identity for the Gowers U3 norm>. A set $A$ of density $\delta$ is <quadratically uniform> when its <balanced subset indicator> $1_A-\delta$ satisfies that condition, not when its unbalanced indicator does. Some definitions bound $\|f\|_{U^3}$ rather than its eighth power; that norm parameter is $\alpha^{1/8}$ in the convention here. Quadratic uniformity controls three-dimensional additive cubes and four-term progression counts; small ordinary linear Fourier coefficients alone are not the same condition.
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