 Anticommuting Frobenius isogenies give opposite traces

If an isogeny of elliptic curves $\pi :E_1\to E_2$ satisfies $\pi {\varphi }_1=-{\varphi }_2\pi$ for their Frobenius isogenies, then $\pi (1-{\varphi }_1)=(1+{\varphi }_2)\pi$. Multiplicativity of the degree of an isogeny gives $\mathrm{deg}(1-{\varphi }_1)=\mathrm{deg}(1+{\varphi }_2)$. The degree parallelogram law then gives the displayed point-count identity, since $\mathrm{deg}(1-{\varphi }_i)=\#E_i({\mathbb{F}}_q)$ and $\mathrm{deg}{\varphi }_i=q$.