lean_workbook_plus_10981

⚠️ Retired — specification defect

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

By taking advantage of the fact that the problem only has one answer, there is an interesting alternate solution: We have $p^2+p=q^2+q+242$ , or $\frac{p\left(p+1\right)}{2}=\frac{q\left(q+1\right)}{2}+121$ . Clearly $p>q$ (otherwise $f\left(q\right)+242=f\left(p\right)\leq f\left(q\right)$ ). But notice then that this is

or

We can try small stuff and realize that $61+60=121$ , so $p=61$ , $q=59$ is a solution. But then there's only one solution, so $\left(61,59\right)$ is the answer.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_10981  (p q : ℕ)
  (f : ℕ → ℕ)
  (h₀ : ∀ x, f x = (x * (x + 1)) / 2)
  (h₁ : 0 < p ∧ 0 < q)
  (h₂ : p^2 + p = q^2 + q + 242)
  (h₃ : f p = f q + 242) :
  p = 61 ∧ q = 59   :=  by sorry

The extra242 difference is inconsistent; the excerpt’s uniqueness claim also requires omitted restrictions.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

Proposed corrected statement

Change the triangular-number difference to121 and state the full positive solution set{(121,120),(61,59),(16,5)}. If the original problem requires p,q prime, restore that condition to isolate(61,59).

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.