First, we show it’s sufficient for \(a\) to have \(Q(A, aA) \leq |A|^2 + \frac{|A|^2 (|A|^2 - 1)}{q-1}\). We have \(|A + a A| \geq \frac{|A|^2 |a A|^2}{Q(A, aQ)} \geq \frac{|A|^4}{|A|^2 + \frac{|A|^2 (|A|^2 - 1)}{q-1}}\). We need to show \(\frac{x^2}{x + \frac{x(x-1)}{q-1}} \geq \frac{\min (x, q)}2\). If \(x {\lt} q\) then

Otherwise, if \(q \leq x\), we need to show

By directly expanding, we have

We have \(2 \leq q\), so this value is nonnegative. Now to show that for some \(a\), \(Q(A, aA) \leq |A|^2 + \frac{|A|^2 (|A|^2 - 1)}{q-1}\). Now we show it suffices to show that \(\sum _{a \neq 0} Q(A, a A) \leq |A|^2 (q-1) + |A|^2 (|A|^2 - 1)\). This is because if all values were larger than \(|A|^2 + \frac{|A|^2 (|A|^2 - 1)}{q-1}\), the sum couldn’t’ve been so small.

To show \(\sum _{a \neq 0} Q(A, a A) \leq |A|^2 (q-1) + |A|^2 (|A|^2 - 1)\), we can use 1.8. The quadruples with \(a = c, b = d\) contribute at most \(|A|^2 (q-1)\), and the other quadruples contribute at most \(|A|^2 (|A|^2 - 1)\), because they determine a unique \(a\).

If \(|A| \leq 1\) then \(|3 A^2 - 3 A^2| = |A|\), and this is greater than \(|A|^2 / 2\). Otherwise, we split into cases by whether \(\frac{A - A}{A - A}\) is the entire universe. If it is, then by 2.1 we have some value \(v = (a-b)/(c-d)\) such that \(|A + vA| \geq \frac{\min (|A|^2, p)}2\). By 1.2, we have \(|A + vA| = |(c-d) A + (a-b) A|\). Now \((c-d) A + (a-b) A\), by 1.5, this is a subset of \(c A + a A - d A - b A\), which is a subset of \(2 A^2 - 2 A^2\), and then \(|3A^2 - 3A^2| = |A^2 - A^2 + 2 A^2 - 2 A^2| \geq |2 A^2 - 2 A^2| \geq \frac{\min (|A|^2, p)}2\).

Otherwise, there must be some value such that \(v = (a-b)/(c-d)\) such that \((a-b+c-d)/(c-d) = (a-b)/(c-d)+1 \notin \frac{A - A}{A - A}\). Because of that \(|A + (a-b+c-d)/(c-d) A| = |A|^2\), so \(|(c-d) A + (a-b+c-d) A| = |A|^2\). Using 1.5 and 1.4 we have \((c-d) A + (a-b+c-d) A \subseteq 3A^2 - 3A^2\), so \(|3A^2 - 3A^2| \geq |A|^2 \geq \frac{|A|^2}2\).