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The $q$-power Frobenius isogeny sends affine coordinates to $(x^q,y^q)$, has degree $q$, and fixes exactly the points defined over ${\mathbb{F}}_q$, including the origin. ConsequentlyThe degree is $q$ because the $q$-power Frobenius on a smooth curve has function-field inseparable degree $q$. By the permitted separability of $1-\varphi$, this separable isogeny has exactly $\mathrm{deg}(1-\varphi )$ geometric kernel points; there is no inseparable multiplicity to remove. Part(i) therefore givesThis relates the elliptic-curve point count over a finite field to the integer trace, rather than treating the point count as an independently assumed definition of the trace.

For every prime $p\ne 37$, the elliptic-curve discriminant is a unit modulo $p$, giving good reduction. Hasse's theorem on elliptic curves givesEquivalently, the elliptic-curve point count over a finite field lies between $p+1-2\sqrt{p}$ and $p+1+2\sqrt{p}$, including for $p=2$.

Let $t=q+1-\#E({\mathbb{F}}_q)$. The stated characteristic equation for the Frobenius isogeny isTake $\alpha ,\beta$ to be the complex roots of $T^2-tT+q$. They satisfy $\alpha +\beta =t$ and $\alpha \beta =q$. By the Hasse bound, $t^2\le 4q$. If the inequality is strict, the roots are complex conjugates and have squared modulus $q$. In the equality case both roots are the same real number $\pm \sqrt{q}$. Thus in every caseTo obtain every extension-field count without assuming distinct roots, set $t_n=\mathrm{tr}({\pi }^n)$ and $t_0=2$. The dual isogeny satisfies $\hat{\pi }\pi =[q]$ and $\pi +\hat{\pi }=[t]$. Consequently the sums ${\pi }^n+{\hat{\pi }}^n=[t_n]$ satisfyAlternatively, apply additivity of the trace of an elliptic-curve endomorphism directly to the characteristic equation multiplied by ${\pi }^{n-2}$. The sequence ${\alpha }^n+{\beta }^n$ has the same recurrence and initial values, so $t_n={\alpha }^n+{\beta }^n$ by induction. This is the Frobenius trace recurrence, valid also at a repeated root.

The fixed points of ${\pi }^n$ are $E({\mathbb{F}}_{q^n})$, and $1-{\pi }^n$ is again separable. Since $\mathrm{deg}{\pi }^n=q^n$, the trace-degree identity givesFactoring and using $\beta =q/\alpha$ yieldsThis is the elliptic-curve point count over a finite field formula and the Riemann hypothesis for an elliptic curve over a finite field. In particular the powers on both factors are necessary; they are damaged in the converted TeX but present in the original PDF.

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.

For this integral Weierstrass equation of an elliptic curve, computeIf an odd prime divides $m$ or $m^2-1$, then $c_4$ is a unit there and $\Delta$ is not. The j-invariant of an elliptic curve consequently has negative valuation. Since good reduction implies an integral j-invariant, such a prime cannot be a prime of good reduction, regardless of whether the displayed equation is minimal. Conversely, if it divides neither factor, this equation already has unit discriminant.

At five the good possibilities have $m\equiv \pm 2$, so $m^2\equiv 4$. At seven they have $m^2\equiv 2$ or $4$. Direct elliptic-curve point count over a finite field gives the following nonzero ordinates, together with the three zero-ordinate points and $O$:At either prime, torsion-freeness of the formal group over Qp for odd p makes reduction injective on all of $E(\mathbb{Q}{)}_{\mathrm{tors}}$, so its order is at most eight. Using only prime-to-$p$ injectivity here would leave an unjustified possible $p$-primary component.

All three nonzero 2-torsion points $(0,0),(-1,0),(-m^2,0)$ are rational. The rational point $P=(m,m(m+1))$ lies on the curve, since $m(m+1)(m+m^2)=m^2(m+1{)}^2$. For an equation $y^2=x(x^2+ax+b)$ the elliptic-curve addition formula givesHere $a=1+m^2$, $b=m^2$, so $2P=(0,0)$ and $P$ has order four. The point $(-1,0)$ is not in $⟨P⟩$, whose only nonzero point of order two is $(0,0)$. Therefore $P$ and $(-1,0)$ generate a subgroup of order eight isomorphic to $\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. The upper bound proves the rational torsion in the family x times x plus one times x plus m squared:

For $x=0,1,2$, the value of $x^3-x-1$ is the nonsquare $2$ in ${\mathbb{F}}_3$. Thus the point at infinity is the only rational point andThe eigenvalues of the Frobenius isogeny are the rootsof $T^2-3T+3$. The elliptic-curve point count over a finite field givesThe last term vanishes exactly when $r\pi /6\equiv \pi /2(\mathrm{mod}\pi )$. Hence

For $E:y^2=x^3+3$, direct counting givesNeither group order is divisible by $17$, so neither group contains a point of order $17$.

At $p=17$ the Frobenius trace is zero. The elliptic-curve point count over a finite field has trace recurrenceFor every $n\ge 1$, this order is congruent to one modulo $17$. Consequently $E({\mathbb{F}}_{17^n})$ has no point of order $17$ for any $n$.

At $p=31$, the trace is $a=31+1-43=-11$. On $E[17]$, Frobenius has characteristic polynomialIts discriminant is $14$, a nonsquare in ${\mathbb{F}}_{17}$, so its two distinct eigenvalues lie in ${\mathbb{F}}_{17^2}^{\times }$. Their orders divide $17^2-1=288$, whence ${\pi }^{288}=1$ on $E[17]$. Thus all of $E[17]$ is rational over ${\mathbb{F}}_{31^{288}}$, and in particular a point of order $17$ exists over some extension with $n\ge 2$.

The zeta function of an elliptic curve over a finite field is the formal power seriesThe proof of Hasse's theorem gives the characteristic equation ${\pi }^2-[a]\pi +[p]=0$. If $\alpha ,\beta$ are the roots of $X^2-aX+p$, then the elliptic-curve point count over a finite field isUsing $\exp ({\sum }_{r\ge 1}{u}^r{T}^r/r)=(1-uT{)}^{-1}$ therefore gives

Let $\pi$ be the Frobenius isogeny of an elliptic curve and putPart (a) givesLet $\alpha ,\beta$ be the two roots of $T^2-aT+p$. The points over ${\mathbb{F}}_{p^r}$ are exactly $\ker (1-{\pi }^r)$. Since the differential of $1-{\pi }^r$ is the identity, this isogeny is separable, and thereforeUsing $\mathrm{deg}\psi =(1-\psi )(1-\hat{\psi })$ in the endomorphism algebra gives the elliptic-curve point count over a finite fieldEquivalently, if $a_r={\alpha }^r+{\beta }^r$, thenand $\#E({\mathbb{F}}_{p^r})=p^r+1-a_r$.