= Solution
The <Hasse theorem for elliptic curves> states
$$
\boxed{\left|\#E(\mathbb F_q)-(q+1)\right|\le2\sqrt q.}
$$
We prove it by the positive <degree of an isogeny>. Let $\pi$ be the $q$-power <Frobenius isogeny>, of degree $q$. Its fixed points are precisely $E(\mathbb F_q)$, so they form the kernel of $1-\pi$. Since the differential of $\pi$ is zero, the differential of $1-\pi$ is the identity; thus $1-\pi$ is a <separable isogeny>. Its degree is the number of its geometric kernel points. Therefore
$$
N:=\#E(\mathbb F_q)=\deg(1-\pi).
$$
We use the standard <degree parallelogram law>, together with $\deg[m]=m^2$: degree, extended by zero at the zero endomorphism, is a positive <quadratic form> on the <endomorphism ring of an elliptic curve>. Put $t=1+q-N$, the <Trace of Frobenius>. Polarizing the degree form gives, for every pair of integers $m,n$,
$$
\deg([m]-[n]\pi)=m^2-tmn+qn^2\ge0.
$$
Indeed, the values on $1$ and $\pi$ are $1$ and $q$, and the cross coefficient is fixed by $\deg(1-\pi)=1+q-t$. These are standard degree facts about <isogenies of elliptic curves>, which suffice without an assumption about characteristic or ordinariness.
For $n\ne0$, division by $n^2$ shows $x^2-tx+q\ge0$ at every rational $x=m/n$. Density of the <rational numbers> and continuity of this polynomial make it nonnegative on the real line. Evaluating at its minimum $x=t/2$ gives $q-t^2/4\ge0$, whence $|t|\le2\sqrt q$. Since $N=q+1-t$, this proves the estimate. This is the <degree-form proof of the Hasse bound>; it proves both sides simultaneously by positivity of one degree form.
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