MathDB

5

Part of 1995 Bulgaria National Olympiad

Problems(1)

exists g(g(a)) = f(a) if f(m) = 1, f(m+n) = m+1 and f(i) = i+1 for all other i

Source: 1995 Bulgaria NMO, Round 4, p5

7/30/2021
Let A={1,2,...,m+n}A = \{1,2,...,m + n\}, where m,nm,n are positive integers, and let the function f : A→AA \to A be defined by: f(m)=1f(m) = 1, f(m+n)=m+1f(m+n) = m+1 and f(i)=i+1f(i) = i+1 for all the other ii. (a) Prove that if mm and nn are odd, then there exists a function g:A→Ag : A \to A such that g(g(a))=f(a)g(g(a)) = f(a) for all a∈Aa \in A. (b) Prove that if mm is even, then there is a function g:A→Ag : A\to A such that g(g(a))=f(a)g(g(a))=f(a) for all a∈Aa \in A is and only if n=mn = m.
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