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I need to prove the following statement:

"Let 'f' and 'g' be two multiplicative functions such that f(pk) = g(pk) for each prime p and k $\geqslant$ 1. Prove that f = g"

I have tried to approach this question by applying it for a particular number n with its prime factorization but couldn't figure how to even proceed.

Any help would be appreciated.

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  • $\begingroup$ Well, what does "multiplicative" mean to you? $\endgroup$ – lulu Feb 18 '18 at 15:42
  • $\begingroup$ f(mn) = f(m)f(n) when m and n are coprime $\endgroup$ – saisanjeev Feb 18 '18 at 15:44
  • $\begingroup$ Ok...so if $n=\prod p_i^{a_i}$ what is $f(n)$? What is $g(n)$? $\endgroup$ – lulu Feb 18 '18 at 15:44
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Yes. Use the prime factorization of $n$ $$ n=p_1^{m_1}\cdots p_k^{m_k} $$ and proceed by recurrence over $k$.

Practically, if $k=0$, you have $n=1$ and the multiplicative functions should coincide to $1$. Having supposed that, for length $k$, they coincide, suppose now $n=p_1^{m_1}\cdots p_{k+1}^{m_{k+1}}$ and set $n'=p_1^{m_1}\cdots p_k^{m_k}$ (the product of the $k$ first terms) so that the last factor ($p_{k+1}^{m_{k+1}}$) is coprime with $n'$. By hypothesis we have $f(n')=g(n')$ and then $$ f(n)=f(n')f(p_{k+1}^{m_{k+1}})=g(n')g(p_{k+1}^{m_{k+1}})=g(n)\ . $$

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  • $\begingroup$ How do prove the functions themselves to be equal? $\endgroup$ – saisanjeev Feb 19 '18 at 5:16
  • $\begingroup$ @saisanjeev I will put more details $\endgroup$ – Duchamp Gérard H. E. Feb 19 '18 at 5:20
  • $\begingroup$ ohh induction.. thanks a lot @duchamp $\endgroup$ – saisanjeev Feb 19 '18 at 5:33

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