Let Z be the integers. Prove that if f : Z → Z satisifies f( f(n) ) = f( f(n+2) + 2 ) = n for all n, and f(0) = 1, then f(n) = 1 - n.
Solution
Easy.
We are given that f(0) = 1 - 0 = 1. Hence f(1) = f( f(0) ) = 0. Suppose f(r) = 1 - r for all r ≤ n. Then f( -(n-2) ) = f( f(n-1) ) = n-1. Hence f(n+1) = f( n-1 + 2) = f( f(-(n-2)) + 2) = -(n-2) - 2 = 1 - (n+1). So the result is true for all positive n. Hence f(-n) = f( f(n+1) ) = n+1 = 1 - (-n), so the result is true for negative n also.
© John Scholes
jscholes@kalva.demon.co.uk
24 Dec 1998