= Solution
The <degree of an isogeny> $\phi:E\to E'$ is the degree of the induced finite extension of function fields. Saying that degree is a quadratic form on $\operatorname{End}(E)$ means
$$
\deg(n\phi)=n^2\deg\phi
$$
and that
$$
\langle\phi,\psi\rangle=\frac12\bigl(\deg(\phi+\psi)-\deg\phi-\deg\psi\bigr)
$$
is bilinear, equivalently
$$
\deg(\phi+\psi)+\deg(\phi-\psi)=2\deg\phi+2\deg\psi.
$$
The <trace of an elliptic-curve endomorphism> is
$$
\operatorname{tr}(\phi)=1+\deg\phi-\deg(1-\phi).
$$
The relation $\phi^2-[\operatorname{tr}\phi]\phi+[\deg\phi]=0$ implies
$$
\operatorname{tr}(\phi^2)=\operatorname{tr}(\phi)^2-2\deg\phi,
$$
the <trace of the square of an elliptic-curve endomorphism>.
The <Hasse theorem for elliptic curves> states that for $E/\mathbb F_q$,
$$
\left|\#E(\mathbb F_q)-(q+1)\right|\leq2\sqrt q.
$$
Let $\pi$ be the <Frobenius isogeny of an elliptic curve>, put $a=\operatorname{tr}(\pi)$, and note that $\deg\pi=q$ and
$$
\#E(\mathbb F_q)=\deg(1-\pi)=q+1-a.
$$
For integers $m,n$, quadraticity gives
$$
0\leq\deg(m+n\pi)=m^2+amn+qn^2.
$$
This binary quadratic form cannot have positive discriminant, since rational numbers $m/n$ are dense, so $a^2-4q\leq0$. Substitution proves the bound. This is the <degree-form proof of the Hasse bound>.
Both endpoints occur. The curve $E:y^2=x^3-x$ over $\mathbb F_3$ is supersingular with trace zero. Over $\mathbb F_9$, its Frobenius is $[-3]$, so
$$
\#E(\mathbb F_9)=9+1-(-6)=16=9+1+2\sqrt9.
$$
Its nontrivial <quadratic twist> over $\mathbb F_9$ has the opposite trace and therefore has
$$
9+1-6=4=9+1-2\sqrt9
$$
points.
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