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I've tried substitution with $0$ and $1$ but it hasn't worked:

Solve for $f(m,n)$:

  1. $m * f(m,np) = f(m,n)f(m,p)$
  2. $f(mn,p) = m* f(n,(p^m))$
  3. $f(m+1,n) = (f(m,n)+ (f(1,n))^m) * n$

Can I have a detailed solution please. I am really scratching my head over this.

Sultan of Quizikhstan
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1 Answers1

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I will assume that $f$ maps $\mathbb N$ into itself. Put $m=1$ in 1) to get $f(1,np)=f(1,n)f(1,p)$. Let us write $g(n)$ for $f(1,n)$. We have $g(np)=g(n)g(p)$. Now put $n=1$ in 2) to get $f(m,p)=mf(1,p^{m})=mg(p^{m})$. This is the form of f (wherein $g$ is an arbitrary multiplicative map of $\mathbb N$). Luckily any function of the form $f(m,p)=mg(p^{m})$ automatically satisfies 3) so the answer is $f(m,p)=mg(p^{m})$ where $g$ is any multiplicative map on $\mathbb N$.

Kavi Rama Murthy
  • 311,013