Are there only finitely many solutions to $i \prod (m _{i} 2 m _{i}) = j \prod (n _{j} 2 n _{j})$ with the $m_{i}, n_{j}$ distinct?

unreviewedOpen. Worked here; no verified result yet.

number theory · disproved · formalized (Lean) · 0 attempts

Check This Result Yourself

use this data

vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

Evidence

unverified AI candidates (1)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

No. In fact there are **infinitely many** such identities with all indices distinct.

candidate solution ↗

Formal proof

AMS 11 · solved (literature)

theorem erdos_397 :
    answer(False) ↔
      {(M, N) : Finset ℕ × Finset ℕ | Disjoint M N ∧
      ∏ i ∈ M, centralBinom i = ∏ j ∈ N, centralBinom j}.Finite

formal-conjectures/397.lean ↗

Connections1

MathOverflow question · discussion

Check it yourself

One command re-derives this record's receipts on your machine.

vela reproduce examples/erdos-problems

Search Vela