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Part of 1997 Bosnia and Herzegovina Team Selection Test

Problems(1)

Bosnia and Herzegovina TST 1997 Day 1 Problem 3

Source: Bosnia and Herzegovina Team Selection Test 1997

9/20/2018
It is given function f:A→Rf : A \rightarrow \mathbb{R}, (A⊆R)(A\subseteq \mathbb{R}) such that f(x+y)=f(x)⋅f(y)−f(xy)+1;(∀x,y∈A)f(x+y)=f(x)\cdot f(y)-f(xy)+1; (\forall x,y \in A) If f:A→Rf : A \rightarrow \mathbb{R}, (N⊆A⊆R)(\mathbb{N} \subseteq A\subseteq \mathbb{R}) is solution of given functional equation, prove that: f(n)={cn+1−1c−1, ∀n∈N,c≠1n+1, ∀n∈N,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)
functionalgebrafunctional equation