Let $f$, $g$ be multiplicative functions, not identically $0$ such that $f(p^k)=g(p^k)$ for each prime $p$ and $k \ge 1$. Prove that $f = g$.

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.

• 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.) Feb 27 '21 at 7:06
• now it's correct. thanks for the edit @AryamanMaithani Feb 27 '21 at 7:13

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 $$f \neq g$$.

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.

• so you're kind of recommending induction, right @Eric Towers ? Feb 27 '21 at 7:08
• @arnavde1220 : Yes. Pretty much any proof using multiplicativity or using unique factorization will end up using an induction explicitly or in disguise. Feb 27 '21 at 7:09

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,

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

• suppose it was not mentioned that f and g are non-zero then this question would be impossible to solve, wouldn't it ? Feb 27 '21 at 7:21
• @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 $f = g$. Feb 27 '21 at 7:30
• 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 $\Bbb N^{>1}$. Feb 27 '21 at 8:09