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Problem 1069

AMC 12 late, AIME early

Algebra Difficulty 5.0 Prove it China Mathematical Competition (Hainan) · China

Let $f$ : $\mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(0) = 1$ and for any $x$ , $y \in \mathbb{R}$ , $f(xy+1) = f(x)f(y) - f(y) - x + 2$ holds. Then $f(x) = \underline{\hspace{2cm}}$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Since for any $x$ , $y \in \mathbb{R}$ , $f(xy+1) = f(x)f(y) - f(y) - x + 2$ , we have
$f(yx + 1) = f(y)f(x) - f(x) - y + 2.$
Thus,
$f(x)f(y) - f(y) - x + 2 = f(y)f(x) - f(x) - y + 2,$
that is,

Put $y = 0$ , we obtain $f(x) = x + 1$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.