lean_workbook_plus_21192

⚠️ 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.

If $x\in[n,n+1)$ , we have $\left[x-1\right]=n-1$ , and we must solve the equation $7\cdotp2018^{n-1}=x^{3}+2x^{2}+3x+1$ , and on the basis of $x\in[n.n+1)$ , we have $(n+1)^{3}+2(n+1)^{2}+3(n+1)+1>7\cdotp2018^{n-1}$ $\geq n^{3}+2n^{2}+3n+1$ . We can see, that if $n>1,$ the LHS inequality is impossible, and if $n<1$ the RHS inequality is inpossible. So we must have $n=1$ thus $\left[x\right]=1$ , and $7=x^{3}+2x^{2}+3x+1$ $\Leftrightarrow x^{3}+2x^{2}+3x-6=0$ $\Leftrightarrow(x-1)(x^{2}+2x+3)=0$ must have a root in the interval $[1,2)$ but this is true for $x=1$ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_21192  (x : ℝ)
  (n : ℤ)
  (h₀ : n ≤ x)
  (h₁ : x < n + 1)
  (h₂ : ⌊x⌋ = n - 1)
  (h₃ : 7 * 2018^(n - 1) = x^3 + 2 * x^2 + 3 * x + 1) :
  n = 1 ∧ ⌊x⌋ = 1 ∧ 7 * 2018^(n - 1) = x^3 + 2 * x^2 + 3 * x + 1   :=  by sorry

The floor typo makes the formal hypotheses inconsistent; the source’s claim about all n<1 and its displayed factorization also require correction.

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

Replace floor(x)=n−1 by floor(x−1)=n−1. Recover the original domain of x before concluding n=1: on unrestricted reals the polynomial equation also has a root in(−1,0) for 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.