Let n be a positive integer, and let C_G(Z_n) denote the number of isomorphism classes of generalized Weierstrass elliptic curves over Z_n. Generalized Weierstrass classification problem. Determine C_G(Z_n). The source states that this number is unknown, so this is posed as an open problem rather than an asserted formula. The paper gives the number of curves in an isomorphism class but leaves the classification count unresolved.

The classification problem for generalized Weierstrass elliptic curves over integer residue rings

About 3 years old · traced to 2023

Let $n$ be a positive integer, and let $C_G(\mathbb{Z}_n)$ denote the number of isomorphism classes of generalized Weierstrass elliptic curves over $\mathbb{Z}_n$ . Generalized Weierstrass classification problem. Determine $C_G(\mathbb{Z}_n)$ .

The source states that this number is unknown, so this is posed as an open problem rather than an asserted formula. The paper gives the number of curves in an isomorphism class but leaves the classification count unresolved.

References

Primary source

Param Parekh, Paavan Parekh, Sourav Deb and Manish K Gupta, “On the Classification of Weierstrass Elliptic Curves over Z_n”, arXiv:2310.11768 (2026).

elliptic-curves modular-arithmetic

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

The paper leaves the count open, but a reader-written argument claims a complete formula, including the difficult cases at the primes two and three, without independent verification.

The problem asks for the number $C_G(\mathbb{Z}_n)$ of isomorphism classes of nonsingular generalized Weierstrass elliptic curves over $\mathbb{Z}_n$ . The 2023 paper explicitly leaves this count as an open problem.

Known results

The number of nonsingular equations is $N_G(\mathbb{Z}_n)=\varphi(n^5)$ , while a fixed class has $\varphi(n^4)/|\operatorname{Aut}(E)|$ representatives.
If $\gcd(n,6)=1$ , generalized and reduced forms have the same classification count.
For $p\ge 5$ , the paper gives explicit prime-power formulas: $2p^m+6$ , $2p^m+2$ , $2p^m+4$ , or $2p^m$ according as $p\equiv1,5,7,11\pmod{12}$ .
Computational values include $C_G(\mathbb{Z}_2)=5$ , $C_G(\mathbb{Z}_3)=8$ , and $C_G(\mathbb{Z}_4)=8$ ; these do not establish a general formula.

Posted attempt

A reader-written argument claims the complete product formula $C_G(\mathbb{Z}_n)=\prod_{p^{e_p}\parallel n}L(p,e_p)$ , with explicit factors for $p=2$ , $p=3$ , and $p\ge5$ , by proving automorphism groups have order $2$ at the exceptional prime powers and using Chinese-remainder factorization. The attempt claims a complete solution but has not been independently verified.

Current status (as of August 2026): The 2023 paper's partial formulas and computations are established, while the full count remains formally open because the reader-written complete formula has no independent verification.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

Let C_g(n) count isomorphism classes of nonsingular generalized Weierstrass equations over Z/nZ. For n=∏p p^{e_p}, the complete answer is C_g(n)=∏{p|n} L(p,e_p), where L(2,1)=5, L(3,1)=8, L(p,e)=2p^e for p∈{2,3}, e≥2, and, for p≥5, L(p,e)=2p^e+c_p, with c_p=6,2,4,0 when p≡1,5,7,11 (mod 12), respectively. The p≥5 factors are the source's proved Theorem 5.5; we resolve its remaining wild prime-power and mixed-composite cases.

Write y²+a₁xy+a₃y=x³+a₂x²+a₄x+a₆. An automorphism (u,r,s,t) obeys ua₁=a₁+2s, u²a₂=a₂−sa₁+3r−s², u³a₃=a₃+ra₁+2t, u⁴a₄=a₄−sa₃+2ra₂−(t+rs)a₁+3r²−2st, u⁶a₆=a₆+ra₄+r²a₂+r³−ta₃−t²−rta₁. The elliptic involution (−1,0,−a₁,−a₃) is always distinct from the identity.

First work modulo 4. Every unit is ±1, so composition with the involution reduces to u=1. The equations simplify to 2s=0, r=sa₁, 2t=−ra₁, sa₃+ta₁=0, ra₄−ta₃−t²−rta₁=0. If a₁ is odd, these imply successively t even and r=s=t=0. If a₁ is even, nonsingularity modulo 2 forces a₃ odd, and the same equations give r=s=t=0. Hence every nonsingular equation modulo 4 has exactly two automorphisms.

Modulo 9, complete the square to obtain y²=x³+Ax²+Bx+C. After composing with the involution, write u=1+3v. The automorphism equations reduce to 3r=6vA, 3vB=2Ar+3r², 0=r(B+Ar+r²). If A is a unit, these force r=v=0. Otherwise nonsingularity forces B to be a unit, and again r=v=0. Thus every nonsingular equation modulo 9 has exactly two automorphisms.

For m≥3, an automorphism reducing to the identity modulo p^{m−1} has u=1+p^{m−1}U, r=p^{m−1}R, s=p^{m−1}S, t=p^{m−1}T. Modulo 2 the linearized equations become a₁U=0, a₁S+R=0, a₃U+a₁R=0, a₃S+a₁T=0, a₄R+a₃T=0. If a₁ is odd they imply U=R=S=T=0; otherwise nonsingularity makes a₃ odd and gives the same conclusion. For p=3 the completed-square equations instead give 2AU=0, BU=2AR, BR=0. Either A is a unit or nonsingularity makes B a unit, so U=R=0. Reduction on automorphism groups is therefore injective at every higher level. Induction from the cases modulo 4 and 9 proves |Aut(E/(Z/p^mZ))|=2 for p=2,3 and every m≥2.

The source's nonsingular-equation count is p^{5m}−p^{5m−1}, while the coordinate-change group has order p^{4m−1}(p−1). Every orbit consequently has half that group order, giving C_g(p^m)=2(p^{5m}−p^{5m−1})/[p^{4m−1}(p−1)]=2p^m for p=2,3 and m≥2. Finally, the Chinese remainder theorem factors both coefficient tuples and coordinate-change groups over prime-power components, so their orbits factor as well. This gives the stated product formula for every n and resolves all cases of the open problem.