lean_workbook_plus_38774

⚠️ 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.

Let $a_n=(1+\sqrt 2)^n +(1-\sqrt 2)^n$ so we can easily get : \n $a_n=2a_{n-1}+a_{n-2} ...(1)$ \n \n Note that if $n$ is odd then $\lfloor (1+\sqrt 2)^n\rfloor=a_n$ \n an if n is even then $\lfloor (1+\sqrt 2)^n\rfloor=a_n-1$ . \n \n so we need to prove that $a_n$ is even $\forall n\in N.$ \n \n from the relation $(1)$ we have $a_n=a_{n-2} mod 2$ and since $a_1=2$ and $a_2=6$ the result follows.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_38774  (a : ℕ → ℝ)
  (h₀ : a 1 = 2)
  (h₁ : a 2 = 6)
  (h₂ : ∀ n, a (n + 2) = 2 * a (n + 1) + a n) :
  ∀ n, Even (Int.floor ((1 + Real.sqrt 2)^n))   :=  by sorry

Evenness of aₙ was replaced by evenness of the floor, which is odd at even indices.

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

Proposed corrected statement

Prove aₙ is even; for the floor state it is even iff n is odd, using floor((1+√2)ⁿ)=aₙ for odd n and aₙ−1 for even n with the zero-index case included.

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.