= Solution
The <Hasse theorem for elliptic curves> states that, for an elliptic curve over $\mathbb F_q$,
$$
\left|\#E(\mathbb F_q)-(q+1)\right|\leq2\sqrt q.
$$
Let $\pi$ be the <Frobenius isogeny of an elliptic curve> and put $a=q+1-\#E(\mathbb F_q)$. The fixed points of $\pi$ are $E(\mathbb F_q)$, and $1-\pi$ is separable, so
$$
\deg(1-\pi)=\#E(\mathbb F_q).
$$
Hence the <trace of an elliptic-curve endomorphism> is
$$
\operatorname{tr}(\pi)=1+\deg\pi-\deg(1-\pi)=a,
$$
while $\deg\pi=q$.
The degree on $\operatorname{End}(E)$ is a nonnegative quadratic form. Polarization and the identities for the <dual isogeny> give, for integers $m,n$,
$$
\deg([m]+[n]\pi)=m^2+amn+qn^2.
$$
If $a^2>4q$, this real binary quadratic form is indefinite. An open cone on which it is negative contains a nonzero rational point and therefore a nonzero integer point, contradicting nonnegativity of isogeny degree. Thus $a^2\leq4q$, which is exactly the claimed inequality.
Back to article page
