PROBLEM 27463, GMB 12/2017, p. 605

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by cip1703, Jan 6, 2018, 1:59 PM

Determine strictly increasing functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $\frac{f(x)+f(y)}{1+f(x+y)}$ is not null natural number, for any $x,y\in \mathbb{N}$ .

(Lucian Dragomir, O\c telu-Ro\c su \c si Nicolae St\u aniloiu, Boc\ sa)

Solution Cip

Answer: $f(x)=ax+1,$ $a\in \mathbb{N}, a\geq 1$

Solution:
For $x=y=0$ we have $\frac{2f(0)}{1+f(0)}\in \mathbb{N}^{\bigstar }$ $\Leftrightarrow 2-\frac{2}{1+f(0)}\in \mathbb{N}^{\bigstar }\Rightarrow f(0)=1.$ The fonction $f$ being strictly increasing, we have $f(x+y)>f(x),f(y)$ for all $x,y\geq 1$ , so $2f(x+y)>f(x)+f(y)$ , then
$0<\frac{f(x)+f(y)}{1+f(x+y)}<\frac{2f(x+y)}{1+f(x+y)}=2-\frac{2}{1+f(x+y)}<2.$ It follows that $\frac{f(x)+f(y)}{1+f(x+y)}=1$ , so
$f(x)+f(y)=f(x+y)+1,$ for all $x,y\geq 1$ .
For $y=1$ we get $f(x+1)-f(x)=f(1)-1$ , and noting $a=f(1)-1$ we have $a> f(0)-1=0$ . From $f(x+1)=f(x)+a$ we get by mathematical induction that $f(x)=ax+1,$ for all $x \in \mathbb{N}$ .
Conversely, all such fonctions satisfies the terms of the statement.
$\raggedleft \blacksquare$

This post has been edited 3 times. Last edited by cip1703, Jan 17, 2018, 6:01 AM
Reason: typo mistake

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PROBLEM 27463, GMB 12/2017, p. 605

by cip1703, Jan 6, 2018, 1:59 PM

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