S_3$. Then, $|S|$ is certainly no more than $M$.

However, we may find a large collection of triples $(x_1, x_2, x_3)$ by choosing $x_1 \in S_1 {\text{ and }}x_2 \in S_2$ arbitrarily, and selecting those for which $x_3 = \frac{-a_1x_1-a_2x_2}{a_3}$ is in $S_3$. If $x_1 = a_2a_3x_1' {\text{ and }}x_2 = a_1a_3x_2'$, then we have $x_3 = -a_1a_2(x_1' + x_2')$. Therefore, a pair $(x_1', x_2')$ will give rise to a solution precisely when $|x_1' + x_2'| {\leqslant}M/6$.

We may compute the number of such pairs $(x_1', x_2')$ as the sum $$\sum_{x_1' = -M/6}^{M/6} M/3 + 1 - |x_1'| = \frac{1}{12}M^2 + O(M).$$ Thus, the number of triples is at least $\frac{1}{12}|S|^2 + O(|S|)$.

Given this, it is natural to define the following quantity:

Define $\gamma_{a_1,a_2,a_3}$ by $$\gamma_{a_1,a_2,a_3} = \limsup_{|S| \rightarrow \infty} \dfrac{T(S)}{|S|^2}$$ where $S$ runs over subsets of ${\mathbb{Z}}$.

Thus, the assertion that $$\label{eqn:gamma_uniform_bound}
\frac{1}{12} {\leqslant}\gamma_{a_1, a_2, a_3} {\leqslant}\frac{3}{4}$$ holds for all $a_1, a_2$ and $a_3$ follows from Proposition \[prop:twelfth\] and the work of Hardy and Littlewood in [@HLIneqNotes].

As far as the author is aware, exact values for $\gamma_{a_1, a_2, a_3}$ are only known in cases for which $|a_1a_2a_3| {\leqslant}2$ (this includes the cases previously discussed). In particular, we 