# Help with multiplicatives

• Apr 2nd 2007, 10:15 AM
lyla
Help with multiplicatives
Can anyone help with this question?

Let f and g be multiplicative functions that are not identically zero and such that f(p^k)=g(p^k) for each prime p and k≥1. Prove that f = g.
• Apr 2nd 2007, 10:20 AM
ThePerfectHacker
Quote:

Originally Posted by lyla
Can anyone help with this question?

Let f and g be multiplicative functions that are not identically zero and such that f(p^k)=g(p^k) for each prime p and k≥1. Prove that f = g.

We want to show that,

f(n)=g(n) for any positive integer n.

Your statement is not actually true :eek: . It is true if n>1.
You need an addition fact that f(1)=g(1).
But anyway, let me show that they agree on n>1.

If n>1 we can prime factorize,
n=p_1^(a_1) * p_2^(a_2) * ... * p_k^(a_k)

Now,
f(n)=f(p_1^(a_1)*p_2^(a_2)*...*p_k^(a_k))=f(p_1^(a _1))*...*f(p_k^(a_k)).
Because,
gcd(p_i,p_j)=1 if i!=j.

The same thing for g(n).
But,
f(p_i^(a_i))=g(p_i^(a_i))
Thus,
f(n)=g(n) for n>1.
• Apr 2nd 2007, 10:29 AM
lyla
Thank you very much, I was confused in seeing when they were equal. :o
• Apr 2nd 2007, 12:06 PM
ThePerfectHacker
Quote:

Originally Posted by lyla
Thank you very much, I was confused in seeing when they were equal. :o

Because,

f(p_1^(a_1))=g(p_1^(a_1))
f(p_2^(a_2))=g(p_2^(a_2))
...
f(p_k^(a_k))=g(p_k^(a_k))

Then multily them out,

f(p_1^(a_1))*...*f(p_k^(a_k))=g(p_1^(a_1))*...*g(p _k^(a_k))
Now you can put them under the same function because it is weakly multiplicative.

f(p_1^(a_1)*...*p_k^(a_K))=g(p_1^(a_1)*...*p_k^(a_ k))

But then,

f(n)=g(n)