## FE FE FE

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

### FE FE FE

We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
$f(x)f(y)=2f(x+yf(x))$
for all positive real numbers $x$ and $y$.
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: FE FE FE

$f(x)f(y+k)=2f(x+(y+k)f(x))=2f(x+yf(x)+\frac{2k}{f(y)}f(x+yf(x))$ [Recall, $f(x)=\frac{2}{f(y)}f(x+yf(x)$]
$\Rightarrow f(x)f(y+k) = f(x+yf(x))f(\frac{2k}{f(y)})=\frac{1}{2}f(x)f(y)f(\frac{2k}{f(y)})$
$\Rightarrow 2f(y+k)=f(y)f(\frac{2k}{f(y)})=2f(y+\frac{2k}{f(y)}f(y))=2f(y+2k)$
Inductivly, $f(y+k)=f(y+nk)$ for all $n \in \mathbb{N}$
So, the function is a constant function.
$\therefore f(x)=2$
This was freedom. Losing all hope was freedom.