2020 CAMO Problems/Problem 1

Problem 1

Let $f:\mathbb R_{>0}\to\mathbb R_{>0}$ (meaning $f$ takes positive real numbers to positive real numbers) be a nonconstant function such that for any positive real numbers $x$ and $y$ , $f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y).$ Prove that there is a constant $a>1$ such that $f(x)=\frac{a^x-1}{a^x+1}$ for all positive real numbers $x$ .

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Because $f(x)f(y)f(x+y)=f(x)+f(y)-f(x+y)$ , we can find that $f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}$ It's obvious that if there exists two real numbers $x$ and $y$ , which satisfies $f(x)=\frac{a^x-1}{a^x+1}$ and $f(y)=\frac{a^y-1}{a^y+1}$

Then, for $f(x+y)$ , $f(x+y)=\frac{f(x)+f(y)}{1+f(x)*f(y)}$ , $f(x+y)=\frac{2*a^(x+y)-2}{2*a^(x+y)+2}$

Then, $f(x+y)=\frac{a^(x+y)-1}{a^(x+y)+1}$

The fraction is also satisfies for $f(x+y)

Then, we can solve this problem using mathematical induction

~~Andy666

2020 CAMO (Problems • Resources)
Preceded by First problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6
All CAMO Problems and Solutions

2020 CJMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6
All CJMO Problems and Solutions

Page

Toolbox

Search

2020 CAMO Problems/Problem 1

Problem 1

Solution

See also