## Functional equation from Japan MO 2016

For discussing Olympiad Level Algebra (and Inequality) problems
Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm
Location: Chittagong, Bangladesh

### Functional equation from Japan MO 2016

Find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$
$$f(yf(x)-x)=f(x)f(y)+2x$$
for all $x,y \in \mathbb{R}$
This was freedom. Losing all hope was freedom.

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm
Location: Chittagong, Bangladesh

### Re: Functional equation from Japan MO 2016

Let $P(x,y)$ denotes the assertion.
$P(0,0) \Rightarrow f(0) = (f(0))^2 \Rightarrow f(0) = 0$ or $1$

Case 1: $f(0)=0$
$P(x,0) \Rightarrow f(-x) = 2x \Rightarrow f(x) = -2x$ which is obviously a solution.

Case 2: $f(0)=1$
$P(x,0) \Rightarrow f(-x) = f(x) + 2x$ ................... (i)
$P(x,-y) \Rightarrow f(-yf(x)-x) = f(x)f(-y) +2x = f(yf(x)+x)+2yf(x)+2x$ where the last equality follows from (i)
$\Rightarrow f(yf(x)+x)=f(x)f(y)$
$\Rightarrow f(yf(x)-x)=f(yf(x)+x)+2x$
$\Rightarrow f(y+x)+2x=f(y-x)$
$\Rightarrow f(y+x)+y+x=f(y-x)+y-x$
$\Rightarrow f(x)+x=c$
$\Rightarrow f(x)=c-x$
If we input $x=0$ we will get $c=1$

All such functions are $-2x$ and $1-x$. It's easy to varify these two.
This was freedom. Losing all hope was freedom.