lean_workbook_plus_6144

⚠️ 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.

It means $ab-b+1=bc-c+1=ca-a+1$. This leads to $b(c-a) = c-b$, $c(a-b) = a-c$, $a(b-c) = b-a$. Multiplying and factoring yields $(a-b)(b-c)(c-a)(abc+1)=0$, thus some two of $a,b,c$ are equal, which immediately forces all three to be equal, thus $a=b=c=x$. Now the only thing needed is $(x^2-x+1)^3 \mid (x^3+1)^2 = (x+1)^2(x^2-x+1)^2$, whence $x^2-x+1 \mid (x+1)^2 = x^2+2x+1$. So $x^2-x+1 \mid 3x$, thus $x^2-x+1 \mid 3$, since $\gcd(x^2-x+1,x)=1$. There are only two possibilities; $x^2-x+1 =1$, forcing $x=1$ (this is excluded in this part), and $x^2-x+1 =3$, forcing $x=2$, thus $\boxed{a=b=c=2}$ is the only other solution.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_6144  (a b c : ℤ)
  (h₀ : 0 < a ∧ 0 < b ∧ 0 < c)
  (h₁ : (a * b - b + 1) * (b * c - c + 1) * (c * a - a + 1) = 1) :
  a = b ∧ b = c ∧ c = 2   :=  by sorry

The source derivation involves divisibility and excludes x=1; the formal premise instead says the product equals 1.

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

Proposed corrected statement

Recover the original divisibility hypotheses and the excluded all-one branch; prove the source classification from those hypotheses.

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.