\[thm:green\_sisask\_structure\], to $S$, with ${\varepsilon}_1 = \left(\frac{{\varepsilon}}{6}\right)^4$. This gives us a decomposition $S = S_1 \amalg \dots \amalg S_n \amalg S_0$. We will start by showing that the contribution to $T(S)$ from solutions $a_1x_1 + a_2x_2 + a_3x_3 = 0$, with at least one of the $x_i$ taken from $S_0$, is small.

\[lem:ignore\_S0\] The number of solutions to $a_1x_1 + a_2x_2 + a_3x_3 = 0$ in $S$, where some $x_i$ is taken from $S_0$, is no greater than $\frac{{\varepsilon}}{2}|S|^2$.

The number of such solutions may be upper bounded by $$T(S_0, S, S) + T(S, S_0, S) + T(S, S, S_0),$$ and so it suffices to show that each term is no greater than $\frac{{\varepsilon}}{6}|S|^2$.

Applying Lemmas \[lem:easy\_TNRG\_lemma\] and \[lem:energy\_cs\], we have $$\begin{aligned}
T(S_0, S, S)^2 &{\leqslant}E(a_1 \cdot S_0, a_2 \cdot S) |S| \\
&{\leqslant}E(S_0, S_0)^{1/2} E(S, S)^{1/2} |S| \\
&{\leqslant}{\varepsilon}_1^{1/2} |S|^4,\end{aligned}$$ from which it follows that $T(S_0, S, S) {\leqslant}{\varepsilon}_1^{1/4}|S|^2$. By our choice of ${\varepsilon}_1$, this gives exactly what we claimed.

At this point, we must bound the number of solutions to $a_1x_1 + a_2x_2 + a_3x_3 = 0$ where each of $x_1, x_2 {\text{ and }}x_3$ is taken from an $S_i$ with $i {\geqslant}1$. To do this, we will 