lean_workbook_plus_68939

⚠️ 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.

Subsolution1Then $f(2)=1$ . Moreover, \n\n $f(p^{2k})f(2) = f(2p^{2k}) = f((1^2+2^2)p^{2k})=f(p^{2k})f(5)$ and so $f(5)=f(2)=1$ . Also, \n\n $f(p^{2k})f(3)=f(3p^{2k}) = f((1^2+3^2)p^{2k}) = f(2)f(5)f(p^{2k})$ and so $f(3)=f(2)f(5)=1$ . Now we have \n\n $f(3)f(5)f(p^{2k})=f(3p^k\cdot 5p^k) = f(34p^{2k})=f(2)f(17p^{2k})=f(2)f((1^2+4^2)p^{2k})=f(2)f(4)f(p^{2k})$ which implies that $f(4)=1$ and $f(17)=1$ . Since $f(4)=1$ , we also have $f(8)=1$ . Now we are in position to prove that $f(n)=1$ for all positive integer $n$ . Suppose that this holds for all $n<t$ where $t\geq 6$ , then for $n=t$ , if $t$ is not a power of prime then it is trivial. So let's suppose that $t=q^{\alpha}$ . If $q=2$ , then when $\alpha=2s+1$ we have that \n\n $f(t) = f(2^s\cdot 2^{s+1})=f(5)f(2^{2s})=1,$ and when $\alpha=2s+2$ we have \n\n $f(t)=f(2^s\cdot 2^{s+2})=f(17)f(2^{2s})=1,$ and so this case is done. If $q$ is odd, then when $p\nmid t-1$ we have \n\n $f(t^2-2t+2)f(p^{2k})=f(t-1)f(p^{2k})$ and so $f(t^2-2t+2)=1$ . If otherwise $p\mid t-1$ , then since $p>5$ , it is true that $2^{v_2(t-1)+2}<t.$ Therefore \n\n $f(t^2-2t+2)f(4)=f(4(t-1))=f(2^{v_2(t-1)+2})f(2^{-(v_2(t-1)+2)}(t-1))=1$ which also implies $f(t^2-2t+2)=1$ . Hence this is true in both cases, and so \n\n $f(4)f(t-2)f(t)=f(4((t-2)^2+t^2))=f(8)f(t^2-2t+2),$ showing that $f(t)=1$ . Thus by induction we know that $f(n)=1$ for all $n$ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_68939  (f : ℕ → ℕ)
  (h₀ : ∀ p : ℕ, Nat.Prime p → ∀ k : ℕ, f (p^k) = 1)
  (h₁ : ∀ k : ℕ, f (2^k) = 1)
  (h₂ : ∀ k : ℕ, f (3^k) = 1)
  (h₃ : ∀ k : ℕ, f (5^k) = 1)
  (h₄ : ∀ k : ℕ, f (17^k) = 1)
  (h₅ : ∀ k : ℕ, f (4^k) = 1)
  (h₆ : ∀ k : ℕ, f (8^k) = 1)
  (h₇ : ∀ n : ℕ, 0 < n → ∃ p : ℕ, p.Prime ∧ p^2 ∣ n) :
  ∀ n : ℕ, 0 < n → f n = 1   :=  by sorry

The full source hypotheses are not supplied by this fragment, so a complete corrected function-classification theorem needs that context.

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

Remove the false assertion that every positive integer has a prime-square divisor; restore the original multiplicativity and functional equations before proving all prime-power cases and then f(n)=1.

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.