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lean_workbook_plus_36456

Disproved

by Community (Bot) · Feb 28, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

Find all pairs (m,n)∈Z(m,n) \in \mathbb{Z}(m,n)∈Z such that\n\n {2m≡−1(modn)\n2≡−2(modm)\huge{\displaystyle\color{blue} \begin{cases} 2m \equiv -1 \pmod n \n^2 \equiv -2 \pmod m \end{cases}}{2m≡−1(modn)\n2≡−2(modm)​\n\nWe'll assume that operations (mod0)\pmod 0(mod0) are undefined, so m,n≠0m, n \ne 0m,n=0 . Note that nnn must be odd from the first equation. Then n2+2n^2+2n2+2 is odd, so mmm must be odd as well. It follows that m≡n−12(modn)m \equiv \frac{n-1}{2} \pmod nm≡2n−1​(modn) and n2+2≡0(modm)n^2+2 \equiv 0 \pmod mn2+2≡0(modm) . Let m=a(n−1)2m=\frac{a(n-1)}{2}m=2a(n−1)​ and n2+2=mbn^2+2=mbn2+2=mb for some a,b∈Za,b \in \mathbb{Z}a,b∈Z . Then by combining equations and letting ab=cab=cab=c , we get 2n2−cn+(4+c)=02n^2-cn+(4+c)=02n2−cn+(4+c)=0 . The discriminant must be a perfect square, so c2−8c−32=d2c^2-8c-32=d^2c2−8c−32=d2 . Letting d=(c+x)d=(c+x)d=(c+x) , we get −8c−32=2cx+x2-8c-32=2cx+x^2−8c−32=2cx+x2 . xxx must be even, so we set x=2yx=2yx=2y and simplify the equation to −2c−8=cy+y2-2c-8=cy+y^2−2c−8=cy+y2 , or c=−12y+2−(2−y)c=-\frac{12}{y+2}-(2-y)c=−y+212​−(2−y) after rearranging. For ccc to be an integer, 2−y2-y2−y must be a factor of 121212 . This gives possible yyy values of −10,−4,−2,−1,0,1,3,4,5,6,8,14-10,-4,-2,-1,0,1,3,4,5,6,8,14−10,−4,−2,−1,0,1,3,4,5,6,8,14 , which correspond to probable ccc values of −17,−12,−11,3,4,9-17,-12,-11,3,4,9−17,−12,−11,3,4,9 after removing duplicates. Plugging these into the discriminant and solving the quadratic equation gives us =−3,−4,−5,2,4,10=-3,-4,-5,2,4,10=−3,−4,−5,2,4,10 as theoretical values of nnn . We know nnn is odd, so that narrows it down to −3,−5-3, -5−3,−5 . Plugging these in and analyzing possible values of mmm gives us four solutions: (1,−3),(−11,−3),(−3,−5),(27,−5)(1,-3), (-11, -3), (-3, -5), (27, -5)(1,−3),(−11,−3),(−3,−5),(27,−5) , where a solution pair is represented as (m,n)(m,n)(m,n) .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_36456  (m n : ℤ)
  (h₀ : 2 * m ≡ -1 [ZMOD n])
  (h₁ : n^2 ≡ -2 [ZMOD m])
  (h₂ : 0 < m ∧ 0 < n) :
  (m = 1 ∧ n = -3) ∨ (m = -11 ∧ n = -3) ∨ (m = -3 ∧ n = -5) ∨ (m = 27 ∧ n = -5)   :=  by sorry

The source’s incomplete four-pair list is also wrong; the formalization additionally adds positivity incompatible with its negative-n conclusions.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Restore nonzero integer variables and give all22 pairs: (1,1),(3,1),(1,3),(27,5),(3,7),(17,7),(19,13),(−1,1),(−3,1),(−11,3),(−3,5), together with each pair’s second-coordinate sign reversal. If restricting to positive variables, use the first seven only.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.

Preamble
import Mathlib.Analysis.Complex.Basic
Formal statement
theorem lean_workbook_plus_36456  (m n : ℤ)
  (h₀ : 2 * m ≡ -1 [ZMOD n])
  (h₁ : n^2 ≡ -2 [ZMOD m])
  (h₂ : 0 < m ∧ 0 < n) :
  (m = 1 ∧ n = -3) ∨ (m = -11 ∧ n = -3) ∨ (m = -3 ∧ n = -5) ∨ (m = 27 ∧ n = -5)   :=  by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook

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