= Anticommuting Frobenius isogenies give opposite traces
{title2=$\#E_1(\mathbb F_q)+\#E_2(\mathbb F_q)=2(q+1)$}
If an <isogeny of elliptic curves> $\pi:E_1\to E_2$ satisfies $\pi\phi_1=-\phi_2\pi$ for their <Frobenius isogenies>, then $\pi(1-\phi_1)=(1+\phi_2)\pi$. Multiplicativity of the <degree of an isogeny> gives $\deg(1-\phi_1)=\deg(1+\phi_2)$. The <degree parallelogram law> then gives the displayed point-count identity, since $\deg(1-\phi_i)=\#E_i(\mathbb F_q)$ and $\deg\phi_i=q$.
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