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Find the function(s) $f: \mathbb{R} \to \mathbb{R}$ which satisfies $f(x)+f(y)-f(x+y)=f(xy)-f(x)f(y)$.

My attempt:

\begin{align} &P(x, 0): f(0)=f(0)-f(x)f(0). \\ &\text{if } f(0) \neq 0: \\ &1=1-f(x). \Rightarrow f \equiv 0, \textbf{ Contradiction.} \\ &\therefore f(0)=0. \\ \ \\ &P(x, 1): f(x)+f(1)-f(x+1)=f(x)-f(x)f(1). \\ &\therefore \big( f(x)+1\big)f(1)=f(x+1). \\ \ \\ &\text{if } f(1)=0: & (1)\\ &f(x+1)=0. \Rightarrow \boxed{f \equiv 0}, \textbf{ Solution.} \\ \ \\ &\text{if } f(1) \neq 0: \\ &P(x, -x): f(x)+f(-x)-f(0)=f(-x^2)-f(x)f(-x). \\ &\Rightarrow f(x)+f(-x)=f(-x^2)-f(x)f(-x). \\ &\Rightarrow \big( f(x)+1 \big) \big( f(-x)+1 \big) =f(-x^2)+1. \\ \ \\ &x=1; \ \big(f(1)+1\big) \big(f(-1)+1 \big) = f(-1)+1. \\ \ \\ &\text{if } f(-1) \neq -1: \\ &f(1)=0. \Rightarrow \textbf{ Contradiction. } \\ \ \\ &\text{if } f(-1) = -1: \\ &P(x, -1): f(x)-1-f(x-1)=f(-x)+f(x). \\ &\therefore f(-x)+1+f(x-1)=0. \end{align}

Mohsen Shahriari
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RDK
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1 Answers1

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This is a partial answer that shows $f(x)=x~~\forall x\in \mathbb{Q}$.
We are considering the case where $f(x)\ne0~~\forall x\ne 0$ .
You have showed that $f(-x)+1+f(x-1)=0$.
Let $x=y+\frac12$
$$\implies f(-y-\frac12)+f(y-\frac12)=-1$$
$$\implies P(-y-\frac12, y-\frac12): -1+1=f(\frac14-y^2)-f(-y-\frac12)f(y-\frac12)$$

Let $z=y-\frac12$ $$\implies f(-z-1)f(z)=f(-z^2-z)$$ Substitutting $z=1$ we get $f(1)=1$ $$\implies f(x+1)=f(x)+1$$ $$\implies f(x+n)=f(x)+n~~~~~~~~~\forall{n\in\mathbb{N}}$$ $$ P(x, n):f(x)+n-(f(x)+n)=f(xn)-nf(x)$$ $$\implies f(xn)=nf(x)~~~~~~~~~\forall{n\in\mathbb{N}}$$ $$\implies f(x)=x~~~~~~~~~\forall{x\in\mathbb{Q}}$$

Asher2211
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