In Silverman's Arithmetic of Elliptic Curves we have the following theorem:
Next it introduces a constant $\alpha$ such that $T(\alpha) = 0$ and makes the following claim:
Now, it should be easy to prove that $T(\beta) = T(\alpha \beta) = 0$. The problem is, expanding any of the two does not wield 0. Any idea on how to show this equality?
Both of the calculations work out for me. I'll show $T(\alpha \beta) = 0$ and leave $T(\beta) = 0$ to you. Note that $T(a) = 2a$ for all $a \in \mathbb{Q}$. For clarity, let $\widetilde{\beta} = \beta - \frac{1}{2} T(\beta) - \frac{T(\alpha \beta)}{2 \alpha^2} \alpha$. Then \begin{align*} \alpha \widetilde{\beta} &= \alpha \left(\beta - \frac{T(\beta)}{2} - \frac{T(\alpha \beta)}{2 \alpha^2} \alpha\right) = \alpha \beta - \frac{T(\beta)}{2} \alpha - \frac{T(\alpha \beta)}{2} \, . \end{align*} Since the trace is $\mathbb{Q}$-linear, then \begin{align*} T(\alpha \widetilde{\beta}) &= T\left(\alpha \beta - \frac{T(\beta)}{2} \alpha - \frac{T(\alpha \beta)}{2}\right) = T(\alpha \beta) - \frac{T(\beta)}{2} \overbrace{T(\alpha)}^0 - T\left(\frac{T(\alpha \beta)}{2}\right)\\ &= T(\alpha \beta) - 2 \cdot \frac{T(\alpha \beta)}{2} = 0 \, . \end{align*}