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Let $f$ and $g$ be multiplicative functions that are not identically $0$ and such that $f(p^k)=g(p^k)$ for each prime $p$ and $k\ge1$. Prove that $f=g$.

Source: Elementary Number Theory by David M. Burton

I just know that any function is multiplicative if $f(mn)=f(m)f(n)$ where $gcd(m,n) = 1$ ,I just need some ideas or hints to solve problems like these.

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    $\begingroup$ Any natural number greater than $1$ can be written as a product of prime powers. Can you conclude from there? (Also, double-check if you've got the correct definition of multiplicative.) $\endgroup$ Feb 27 '21 at 7:06
  • $\begingroup$ now it's correct. thanks for the edit @AryamanMaithani $\endgroup$
    – arnav_de
    Feb 27 '21 at 7:13
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As the other answers have shown, $f(n) = g(n)$ for all $n > 1$ follows from prime factorisation. (You only need existence, not even uniqueness.) There was no use of $f$ and $g$ not being identically zero so far.

However, for $n = 1$, the above argument does not work. Now, we know that $$f(n) = f(n)f(1)$$ for all $n \in \Bbb N$. Since $f$ is not identically $0$, there exists $n \in \Bbb N$ such that $f(n) \neq 0$. From that, we can conclude that $f(1) = 1.$ Similarly, $g(1) = 1$. This lets you conclude that $f = g$.


What if you weren't given $f$ and $g$ are not identically zero?

Well, consider $f \equiv 0$ but $g$ to be defined as $$g(n) = \begin{cases}1 & n = 1\\ 0 & n > 1\end{cases}.$$

Then, $f(p^k) = g(p^k)$ for all primes $p$ and $k \ge 1$ (and both are (completely!) multiplicative) but $f \neq g$.

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This is a sketch. Many details are missing. You should be able to fill those in by following the patterns you are seeing in the various proofs in your source material.

$f = g$ means: for all $n$, $f(n) = g(n)$.

So let $n$ be arbitrary. By unique factorization, $$ n = p_1^{n_1} p_2^{n_2} \cdots p_m^{n_m} \text{,} $$ for some nonnegative integer $m$, a collection of distinct primes, $p_1, \dots, p_m$, and a collection of positive integers $n_1, \dots, n_m$.

You want to show the proposition $$ P_m: f(p_1^{n_1} p_2^{n_2} \cdots p_m^{n_m}) = g(p_1^{n_1} p_2^{n_2} \cdots p_m^{n_m}) \text{.} $$ You are given that it is true when $m = 1$. Can you show that if $P_M$ is true (for some $M \geq 1$), then $P_{M+1}m$ is true using multiplicativity? If you put these two facts together, you have a complete proof.

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  • $\begingroup$ so you're kind of recommending induction, right @Eric Towers ? $\endgroup$
    – arnav_de
    Feb 27 '21 at 7:08
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    $\begingroup$ @arnavde1220 : Yes. Pretty much any proof using multiplicativity or using unique factorization will end up using an induction explicitly or in disguise. $\endgroup$ Feb 27 '21 at 7:09
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Assuming $f $ and $g$ are non-zero functions, for any $m\in \mathbb{N^{>1}}$, let $m=p_1^{\alpha_1}p_2^{\alpha_2}\cdot \cdot \cdot p_k^{\alpha_k}$ be its prime factorization. Then clearly,

$f(m)=f(p_1^{\alpha_1}p_2^{\alpha_2}\cdot \cdot \cdot p_k^{\alpha_k})=f(p_1^{\alpha_1})f(p_2^{\alpha_2})\cdot \cdot f(p_k^{\alpha_k})=g(p_1^{\alpha_1})g(p_2^{\alpha_2})\cdot \cdot g(p_k^{\alpha_k})=g(p_1^{\alpha_1}p_2^{\alpha_2}\cdot \cdot \cdot p_k^{\alpha_k})=g(m).$

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  • $\begingroup$ suppose it was not mentioned that f and g are non-zero then this question would be impossible to solve, wouldn't it ? $\endgroup$
    – arnav_de
    Feb 27 '21 at 7:21
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    $\begingroup$ @arnavde1220 : No. If $f$ is the zero function, then $f(p^k) = 0$ for all $p$ and all $k$, and then $g(p^k) = 0$ for all $p$ and all $k$. This would make $g$ also the zero function. The same argument in reverse shows that if either is the zero function, then both are and $f = g$. $\endgroup$ Feb 27 '21 at 7:30
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    $\begingroup$ I think $f$ and $g$ being not identically zero tells you that $f(1) = 1 = g(1).$ Note that the argument above only tells you that $f$ and $g$ agree on $\Bbb N^{>1}$. $\endgroup$ Feb 27 '21 at 8:09

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