= Solution
Let $\pi$ denote the <Frobenius isogeny of an elliptic curve> and write $N=\#E(\mathbb F_q)$. Its <degree of an isogeny> is $q$. The <isogeny of elliptic curves> $1-\pi$ has differential equal to the identity because $d\pi=0$, so it is a <separable isogeny>. Its kernel consists precisely of the <rational points> fixed by $\pi$, giving
$$
\deg(1-\pi)=N.
$$
We use the <degree parallelogram law> for <isogenies of elliptic curves>, with degree zero assigned to the zero map:
$$
\deg(f+g)+\deg(f-g)=2\deg f+2\deg g,\qquad \deg[n]=n^2.
$$
One explanation of the first identity is the <divisor proof of the degree parallelogram law>: on $E\times E$, the zero <divisor> of $x(P)-x(Q)$ is the sum of the diagonal and the graph of negation, while its pole <divisor> is twice each coordinate copy of $O$. Pulling the associated <line bundle> identity back by $(f,g)$ and taking degrees gives the identity, including exceptional cases by the line-bundle formulation. This works for a general <Weierstrass equation of an elliptic curve>, including characteristic two; $x$ is the quotient coordinate for negation. Polarization therefore makes degree a <quadratic form> on the <endomorphism ring of an elliptic curve>.
Set $a=q+1-N$, the <Trace of Frobenius>. The cross term is determined by $\deg(1-\pi)=1+q-a$, giving, for all integers $m,n$,
$$
\deg([m]-[n]\pi)=m^2-amn+qn^2\geq0.
$$
If $a^2>4q$, the real degree-two <polynomial> $X^2-aX+q$ is negative on a nonempty open interval. That interval contains a rational $m/n$, contradicting the displayed nonnegativity after multiplication by $n^2$. Thus $a^2\leq4q$. \b[Hasse's bounds are]
$$
\boxed{q+1-2\sqrt q\ \leq\#E(\mathbb F_q)\ \leq q+1+2\sqrt q.}
$$
The argument proves the <Hasse theorem for elliptic curves> without assuming its bound in advance.
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