The conjecture on the values of n for which M(n)=2

From papers

Let M(n)M(n) denote the minimum number of distinct positive integers in an nn-tuple whose second elementary symmetric sum equals its nnth elementary symmetric sum. For nN3n\in\mathbb{N}_{\geq 3}, consider the values of nn for which M(n)=2M(n)=2.

Conjecture. The only values of nN3n\in\mathbb{N}_{\geq 3} such that M(n)=2M(n)=2 are

n=3,4,5,11,41156.n=3,4,5,11,41156.

In particular,

lim infn+M(n)=3.\liminf_{n\rightarrow+\infty}M(n)=3.

This is motivated by the reduction of the case M(n)=2M(n)=2 to integral points on the curves y2=8xk+4kx24kx+1y^2=8x^k+4kx^2-4kx+1, together with explicit analyses for small kk and the fact that the associated curves have genus at least 22 for k5k\geq 5. The asserted absence of further integral points for larger kk is based on numerical calculations and remains open.

Progress summary

Open

The proposed list remains an unproved conjecture, with no verified new examples or proof found.

The conjecture claims that exactly n=3,4,5,11,41156n=3,4,5,11,41156 satisfy M(n)=2M(n)=2, and therefore that the minimum number of distinct entries eventually has lower limit 33. It was stated in a 2022 paper, where the final exclusions were explicitly left conjectural.

Known results

  • The case M(n)=2M(n)=2 reduces to integral points on y2=8xk+4kx24kx+1y^2=8x^k+4kx^2-4kx+1.
  • Explicit analysis for k=3k=3 and k=4k=4 yields n=3,4,5,11,41156n=3,4,5,11,41156.
  • For k5k\geq5, the associated curves have genus g2g\geq2; numerical calculations suggest no further relevant integral points, but do not prove this.
  • A 2026 paper treats broader elementary-symmetric equations but reports no proof or counterexample for this specific conjecture.

Current status (as of August 2026): The reduction and the explicitly analyzed cases are established, but completeness of the list and the resulting claim about lim infn+M(n)\liminf_{n\to+\infty}M(n) remain open.

Sources
Sources & referencesView supporting material

Primary source

Piotr Miska and Maciej Ulas, “On the Diophantine equation σ_2(X_n)=σ_n(X_n)”, arXiv:2203.03942 (2022).

Solutions 1

Counterexample

The proposed classification

M(n)=2n{3,4,5,11,41156}M(n)=2\quad\Longleftrightarrow\quad n\in\{3,4,5,11,41156\}

fails in both directions.

First, take the nondecreasing positive integer tuple

X=(1,,143,7,7,7,7).X=(\underbrace{1,\ldots,1}_{43},7,7,7,7).

Its product is

σ47(X)=74=2401,\sigma_{47}(X)=7^4=2401,

and separating pairs according to how many entries equal 77 gives

σ2(X)=(432)+4347+(42)72=903+1204+294=2401.\begin{aligned} \sigma_2(X) &=\binom{43}{2}+43\cdot4\cdot7+\binom42\,7^2\\ &=903+1204+294\\ &=2401. \end{aligned}

Thus XX is a solution with exactly two distinct entries, so M(47)2M(47)\le2. A constant solution of length 4747 would require

(472)t2=t47,hencet45=1081=2347.\binom{47}{2}t^2=t^{47}, \qquad\text{hence}\qquad t^{45}=1081=23\cdot47.

No positive integer satisfies this equation: t=1t=1 gives 11, while t2t\ge2 gives t45245>1081t^{45}\ge2^{45}>1081. Consequently

M(47)=2,M(47)=2,

although 4747 is absent from the proposed list.

Second, the constant tuple (3,3,3)(3,3,3) satisfies

σ2(3,3,3)=332=27=33=σ3(3,3,3),\sigma_2(3,3,3)=3\cdot3^2=27=3^3=\sigma_3(3,3,3),

and therefore

M(3)=12,M(3)=1\ne2,

although 33 appears in the proposed list.

Both underlying examples already occur immediately before the conjecture in its source: the displayed conjectural classification is inconsistent with those preceding results. A corrected completeness statement would be a different problem.

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Shivam Patel ·