Functional equation from Japan MO 2016

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Atonu Roy Chowdhury
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Functional equation from Japan MO 2016

Unread post by Atonu Roy Chowdhury » Thu Mar 15, 2018 1:27 pm

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.

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

Re: Functional equation from Japan MO 2016

Unread post by Atonu Roy Chowdhury » Thu Mar 15, 2018 2:50 pm

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.

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