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Bosnia and Herzegovina TST 1997 Day 1 Problem 3

Source: Bosnia and Herzegovina Team Selection Test 1997

September 20, 2018
functionalgebrafunctional equation

Problem Statement

It is given function f:ARf : A \rightarrow \mathbb{R}, (AR)(A\subseteq \mathbb{R}) such that f(x+y)=f(x)f(y)f(xy)+1;(x,yA)f(x+y)=f(x)\cdot f(y)-f(xy)+1; (\forall x,y \in A) If f:ARf : A \rightarrow \mathbb{R}, (NAR)(\mathbb{N} \subseteq A\subseteq \mathbb{R}) is solution of given functional equation, prove that: f(n)={cn+11c1nN,c1n+1nN,c=1f(n)=\begin{cases} \frac{c^{n+1}-1}{c-1} \text{, } \forall n \in \mathbb{N}, c \neq 1 \\ n+1 \text{, } \forall n \in \mathbb{N}, c = 1 \end{cases} where c=f(1)1c=f(1)-1 a)a) Solve given functional equation for A=NA=\mathbb{N} b)b) With A=QA=\mathbb{Q}, find all functions ff which are solutions of the given functional equation and also f(1997)f(1998)f(1997) \neq f(1998)