lean_workbook_plus_47901

⚠️ 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.

Find all functions $f:\mathbb R\to \mathbb R$ such that $f(xf(y)+y)+f(xy+x)=f(x+y)+kxy$ for all $x,y\in \mathbb R$ and $k$ is some constant.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_47901 (k : ℝ): (∀ (f : ℝ → ℝ), (∀ x y, f (x * f y + y) + f (x * y + x) = f (x + y) + k * x * y) ↔ ∃ c:ℝ, ∀ x, f x = c * x + 1)   :=  by sorry

The zero function for k=0 is already excluded by that false answer; a full corrected family is not established here.

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

Proposed corrected statement

Restore the task of classifying pairs(k,f) satisfying the functional equation and remove the invented affine-intercept-one family.

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.