Find all functions $f: \mathbb{N}^{+}\to \mathbb{N}^{+}$ which satisfy (a) $f(2) =2$ (b) $f$ is [[Multiplicative Functions|multiplicative]] (c) for all $m>n,$ $f(m)>f(n)$ ###### Solution We have $f(1)=1.$ We have $f(3)f(5)=f(15)<f(18)=f(2)f(9)=2f(9)$and $f(9)<f(10)=2f(5)$Combining gives $f(3)f(5)<4f(5)$giving $f(3)<4.$ Since $2=f(2)<f(3)<4,$ we conclude that $f(3)=3.$ Now suppose there exists $n> 2$ (which gives $n(n-1)>n$), such that for all $k\leq n,$ $f(k)=k.$ Since $\gcd(k,k-1)=1,$ $f(n(n-1))=f(n)f(n-1)=n(n-1)$Invoking (c), we gain $n=f(n)<f(n+1)<f(n+2)<\dots <f(n^{2}-n)$which gives $f(n+1) = n+1, f(n+2)=n+2,\dots,f(n(n-1))=n(n-1).$ Thus by [[Induction Principle|induction]] principle $f$ is indeed the identity function on $\mathbb{N}^{+}.$ # References Functional equations: a problem-solving approach.