Suppose fn and gn are multiplicative and that fptau gptau f

Suppose f(n) and g(n) are multiplicative and that f(p^tau) = g(p^tau) for each r and each prime p. Prove that f(n)= g(n) for all n.

Solution

Here a function f is multiplicative if f(mn)=f(m)f(n) for all integers m and n where f(1)= 1 always.

Thus f(p^r) = f(p)f(p)f(p).......r times

and g(p^r) = g(p)g(p)g(p).......r times

So when p be 1, then clearly f(1)=1=g(1)

on generalising this result, we find f(n)= g(n)

Proved