$\sum\limits_{\mathbb{d|n}}{f(d)}=\sum\limits_{\mathbb{d|n}}{g(d)}\implies f(n)=g(n)?$

Question

Question:

Is it true that if for functions $f,g$ which map naturals to naturals

For all natural numbers n, we have $f(n)=g(n) \iff$ for all natural numbers n we have $\sum\limits_{\mathbb{d|n}}{f(d)}=\sum\limits_{\mathbb{d|n}}{g(d)}$?

Do divisor sums have this injective like quality on functions on the natural numbers?

I think there may be some counter example. Note that if we impose that $f(1)=g(1)$ we also get that these functions agree on atleast all primes. And then if we also ask $f,g$ are multiplicative then we are done. So there should be some counterexample if we don't get all of these convenient restrictions. Right?

My Motives:

My interest in divisor sums comes out of exploring the number of integer solutions on hyperspheres. Let $\phi(n,r)$ be the number of integer solutions of $\sum\limits_{i=1}^n x_i^2=r$.

Then $\phi(2,r)=4\sum\limits_{d|r}\chi(d)$ where $\chi (x)=sin(\frac{\pi x}{2})=\cases{ 1\text{ when }x\cong 1 \text{ mod }4 \\ -1 \text{ when }x\cong 3 \text{ mod }4 \\ 0 \text{ when } 2|x }$

And $\phi(4,r)=8\sum\limits_{d|r}\psi(d)$ where $\psi (x)=\cases{x \text{ when }x\ncong 0 \text{ mod }4 \\ 0 \text{ when }x\cong 0 \text{ mod }4 }$

So this explains my motivations in exploring divisor sums.

Bonus Question

As a bonus question I am asking for leads on theta series. What else is known about $\phi(n,r)$? Is there an explicit formula for other $n$? Is there an explicit formula in $n$ and $r$?

Answer 1

It's a straightforward strong induction, using the fact that if $d\mid n$ and $d\not=n$, then $d\lt n$.

To get started, we have

$$f(1)=\sum_{d\mid1}f(d)=\sum_{d\mid1}g(d)=g(1)$$

Now if $f(k)=g(k)$ for all $k\lt n$, then

$$f(n)=\sum_{d\mid n}f(d)-\sum_{d\mid n,d\lt n}f(d)=\sum_{d\mid n}g(d)-\sum_{d\mid n,d\lt n}g(d)=g(n)$$

Answer 2

This is an application of Möbius inversion formula

---

$\Rightarrow$

Because $f(n)=g(n), \forall n$ then $f(d)=g(d)$, where $d \mid n$, in particular and then $$\sum\limits_{d\mid n}f(d)=\sum\limits_{d\mid n}g(d)$$ for $\forall n$

---

$\Leftarrow$

Let's note $F(n)=\sum\limits_{d\mid n}f(d)$ and $G(n)=\sum\limits_{d\mid n}g(d)$ and $F(n)=G(n), \forall n$, then using Möbius inversion formula

$$f(n)=\sum\limits_{d\mid n}\mu(d)F\left(\frac{n}{d}\right)$$ $$g(n)=\sum\limits_{d\mid n}\mu(d)G\left(\frac{n}{d}\right)$$ and because $\mu(d)$ is the same for both expressions and $F\left(\frac{n}{d}\right)=G\left(\frac{n}{d}\right)$ in particular, it follows that $f(n)=g(n), \forall n$.