= Solution
Write $N=\#E(\mathbb F_q)$. The <Hasse bound> is
$$
\boxed{|N-(q+1)|\leq2\sqrt q.}
$$
We prove it using <isogeny degrees> of <endomorphisms>, with no assumption about a pre-existing Frobenius eigenvalue estimate. The standard <elliptic isogeny> facts we use are these: <isogeny degree> is multiplicative under composition; $\deg[m]=m^2$; assigning <isogeny degree> zero to the zero <homomorphism> makes <isogeny degree> a <quadratic form>, satisfying the <degree parallelogram law>; and the <isogeny degree> of a <separable isogeny> equals the number of its geometric <kernel of an isogeny> points.
Let $\phi$ be the $q$-power <Frobenius isogeny>. It has <isogeny degree> $q$ and zero differential. Therefore $1-\phi$ has differential the identity and is a <separable isogeny>. Its <kernel of an isogeny> consists exactly of the points fixed by Frobenius, including the identity, so
$$
\deg(1-\phi)=N.
$$
Put $t=q+1-N$. Polarization of the <isogeny degree> <quadratic form> gives, for all <integers> $m,n$,
$$
\deg([m]-[n]\phi)=m^2-tmn+qn^2.
$$
The cross coefficient is fixed by the case $m=n=1$, where the <isogeny degree> is $N=1-t+q$; the coefficients of $m^2,n^2$ are $\deg1=1$ and $\deg\phi=q$.
Every <isogeny degree> is nonnegative. For $n\ne0$, divide by $n^2$ to obtain $u^2-tu+q\geq0$ for every rational $u=m/n$. Continuity and density of the rationals imply the same inequality for real $u$. Its minimum, attained at $u=t/2$, is $q-t^2/4$, so $t^2\leq4q$. This is precisely the stated bound. Equality is allowed; a zero value of the <quadratic form> can occur when the two <endomorphisms> involved are rationally dependent.
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