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The squares in the finite field ${\mathbb{F}}_7$ are $0,1,2,4$. For $x=0,1,\dots ,6$, the values of $x^3-x+1$ and the numbers of possible $y$ are respectivelyAdding $O$ yields $\#E({\mathbb{F}}_7)=12$. The Trace of Frobenius is $a=8-12=-4$. The trace of the square of an elliptic-curve endomorphism is $a^2-2q=16-14=2$, so the elliptic-curve point count over a finite field givesFor example, this trace identity follows from ${\pi }^2-[a]\pi +[q]=0$ and $\mathrm{tr}[q]=2q$.

One suitable second elliptic curve isHere $4\cdot 3^3+27\cdot 6^2\equiv 2(\mathrm{mod}7)$, so its elliptic-curve discriminant is nonzero. Its complete list of rational points is $O,(3,0),(6,3),(6,4)$, obtained by checking the seven $x$-values. For $P=(6,3)$ the tangent slope is $1$ in ${\mathbb{F}}_7$, and the elliptic-curve addition formula gives $2P=(3,0)$. Consequently $P$ has order four and $E^{\prime }({\mathbb{F}}_7)\cong \mathbb{Z}/4\mathbb{Z}$.

Let $\pi$ be the Frobenius isogeny and let $a=p+1-N$ be its Trace of Frobenius. Its characteristic equation is ${\pi }^2-[a]\pi +[p]=0$, so the trace of the square of an elliptic-curve endomorphism is $a^2-2p$. The elliptic-curve point count over a finite field therefore givesThe prime-square point-count formula is consequentlyOne can also obtain it directly from the degree of an isogeny: the maps $1-\pi$ and $1+\pi$ are separable isogenies, with degrees $p+1-a$ and $p+1+a$. Their composition is $1-{\pi }^2$, whose kernel of an isogeny is exactly $E({\mathbb{F}}_{p^2})$. Multiplicativity of degree gives the same product.

The degree of an isogeny $\varphi :E\to E^{\prime }$ is the degree of the induced finite extension of function fields. Saying that degree is a quadratic form on $\mathrm{End}(E)$ meansand thatis bilinear, equivalentlyThe trace of an elliptic-curve endomorphism isThe relation ${\varphi }^2-[\mathrm{tr}\varphi ]\varphi +[\mathrm{deg}\varphi ]=0$ impliesthe trace of the square of an elliptic-curve endomorphism.

The Hasse theorem for elliptic curves states that for $E/{\mathbb{F}}_q$,Let $\pi$ be the Frobenius isogeny of an elliptic curve, put $a=\mathrm{tr}(\pi )$, and note that $\mathrm{deg}\pi =q$ andFor integers $m,n$, quadraticity givesThis binary quadratic form cannot have positive discriminant, since rational numbers $m/n$ are dense, so $a^2-4q\le 0$. 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 $[-3]$, soIts nontrivial quadratic twist over ${\mathbb{F}}_9$ has the opposite trace and therefore haspoints.

For an elliptic curve over ${\mathbb{F}}_p$, put $N=\#E({\mathbb{F}}_p)$ and $a=p+1-N$. The trace of the square of an elliptic-curve endomorphism gives $\mathrm{tr}({\pi }^2)=a^2-2p$ for the Frobenius isogeny. Therefore $\#E({\mathbb{F}}_{p^2})=p^2+1-a^2+2p=N(2p+2-N)$.