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The nonzero quadratic residues modulo five are $1,4$. The values of $x^3+x+1$ and the numbers of possible $y$ areThere are eight affine points and one point at infinity. Thus $\#E({\mathbb{F}}_5)=9$ and part(ii) gives $\boxed{\mathrm{tr}(\varphi )=5+1-9=-3}$. The curve is nonsingular, since its short-Weierstrass discriminant $-16(4+27)$ is nonzero modulo five.

The dual operation reverses products, so $\widehat{{\varphi }^2}=(\hat{\varphi }{)}^2$. Also $\varphi \hat{\varphi }=[5]$ and $\varphi +\hat{\varphi }=[-3]$. Squaring the latter identity givesTherefore $\boxed{\mathrm{tr}({\varphi }^2)=-1}$. Over ${\mathbb{F}}_{25}$ the Frobenius is ${\varphi }^2$, of degree $25$, soMore generally the Frobenius trace recurrence has $t_0=2$, $t_1=t$ and $t_j=t{t}_{j-1}-q{t}_{j-2}$; the square calculation is its first nontrivial case.

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.