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The elliptic curves $E_1:y^2=x^3-x$ and $E_2:y^2=x^3-x+1$ over ${\mathbb{F}}_5$ both have eight rational points. The first has four rational 2-torsion points and the second has two, so their point groups are not isomorphic. They are nevertheless degree-two isogenous via $X=x+2/(x-1)$, $Y=y(1-2/(x-1{)}^2)$. The map sends $O$ and $(1,0)$ to $O$; its pole divisor for $X$ is $2O+2(1,0)$, giving degree two. Their j-invariant of an elliptic curve values are respectively three and one, so the curves are not geometrically isomorphic either.

For a Short Weierstrass form $y^2=x^3+Ax+B$, use $c_4=-48A$ and $\Delta =-16(4A^3+27B^2)$. Both curves have $A=-1$, so $c_4=48=3$ in ${\mathbb{F}}_5$. For $E_1$, $B=0$ gives ${\Delta }_1=64=4$; for $E_2$, $B=1$ gives ${\Delta }_2=-368=2$. ThusThe j-invariant of an elliptic curve is preserved by every isomorphism over the algebraic closure. Since these values differ, the two curves are not geometrically isomorphic, even though the preceding part gives a degree-two elliptic isogeny.

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 integral $m\ge 2$, suppose $E:y^2=x(x+1)(x+m^2)$ has good reduction at five or seven. At five, $m^2\equiv 4$; at seven, $m^2\equiv 2$ or $4$. Each resulting reduced elliptic curve has eight points. These restrictions follow from $\Delta =16m^4(m^2-1{)}^2$ and $c_4=16(m^4-m^2+1)$: if an odd prime divides $m(m^2-1)$, the j-invariant of an elliptic curve has negative valuation. The torsion-freeness of the formal group over Qp for odd p bounds the whole rational torsion by eight. The point $(m,m(m+1))$ doubles to $(0,0)$, and $(-1,0)$ is an independent 2-torsion point. These generate the stated group of order eight.