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Anyone know of any periodic functions satisfying $f(xy)=f(x)+f(y)$, when gcd(x,y)=1,

I need a function other then the function $a_d(k)$, thats 1 if d divides k, and 0 if it doesn't.

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We see at [enter link description here][1] [1]:… – H.T.H Dec 26 '12 at 1:07
Trivial example: $f(x)=0$ – Amr Dec 26 '12 at 1:29
@Ethan $a_6(6)=1$, but $a_6(2)=0$ and $a_6(3)=0$, so $a_6(6) \neq a_6(2)+a_6(3)$. – Matthew Conroy Dec 26 '12 at 1:45
It seems that the domain is $\mathbb N$. Do you want to add the number-theory and arithmetic-functions tags? – Rahul Dec 26 '12 at 1:47
up vote 1 down vote accepted

Summary: I prove that $a_p$ is all you have for $p$ a prime, the last 2 paragraphs is an actual answer to your question (yes, such functions exist).

The collection $V_p$ of all such functions with period $p$ naturally form a complex vector space, it's reasonable to ask for its dimension. It's certainly no more than $p$ dimensional, so always finite.

Let's classify all additive functions $f: \mathbb{N} \rightarrow \mathbb{C}$ with period 2. Indeed, all our data is two numbers $f(0) = z_0, f(1) = z_1$. We have $f(3 \cdot 5) = f(3) + f(5)$, so we have $z_1 = 2z_1$, so in fact $z_1 = 0$, so $V_2 = \langle a_2 \rangle$.

Similarly, let's classify those additive functions with period $3$. Again we have the data of $z_0, z_1, z_2$. Okay yes, by Dirichlet's theorem on primes, we can find infinitely many, but in particular just 2 distinct primes $q, q'$ satisfying $q, q' \equiv 1 \mod 3$. Hence $f(qq') = z_1 = f(q) + f(q') = 2z_1$, so $z_1 = 0$. Similarly we may take $q, q' \equiv 2 \mod 3$, we have $z_1 = f(qq') = f(q) + f(q'') = 2z_2$ so $z_2 = 0$ as well. Hence $V_3 = \langle a_3 \rangle$.

Okay, yeah in fact I claim $V_p = \langle a_p \rangle$ whenever $p$ is prime. To see this, again we have the data of $z_0 \ldots z_{p-1}$. I claim first that $z_1 = 0$. Indeed, as before, take $q, q' \equiv 1 \mod p$, we have $f(qq') = z_1 = f(q) + f(q') = 2z_1$. Now I claim $z_i = 0$ for $i \geqslant 1$. Indeed, we have $i^n = 1 \mod p$ for some $n$ (say $p-1$), hence take primes $q_1 \ldots q_n \equiv i \mod p$, we have $$z_1 = f(\prod q_i) = \sum f(q_i) = n \cdot z_i$$but $z_1 = 0$. So $V_p = \langle a_p \rangle$.

Indeed, for $V_n$, the same argument above shows $z_i = 0$ for $i \in \mathbb{Z}/n\mathbb{Z}^*$. Moving away from prime periods, we can get things more interesting than $a_p$, say for $p = 4$. Indeed, we have $z_1 = z_3 = 0$, but one can directly check that $z_0$ and $z_2$ can be arbitrary, which gives us our desired examples.

(for the direct check, suppose $(n,m) = 1$, we want $f(nm) = f(n) + f(m)$. For $n,m$ both $\equiv$ 1 or 3, we have $0 = 0 + 0$. For WLOG $n \equiv 1,3$ $m \equiv 0,2$, we have $z_{0,2} = z_{0,2} + 0$, and of course if $n,m$ are both among 0 and 2 $\mod 4$, they aren't coprime!)

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My example fits this as $av_2(x)+bv_3(x)$. Nice general result. +1 – coffeemath Dec 26 '12 at 12:20

$f(x)=\frac{x}{2}$, when x is even,

$f(x)=\frac{(x+1)}{2}$, when x is odd,

$f(xy)=f(x)+f(y)$, when $gcd(x,y)=1$





we also have, $f(6)=3$

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But $f(10)=5$, while $f(2)=1$ and $f(5)=3$, so $f(10)\neq f(2)+f(5)$. Also, your $f$ is non-decreasing; it is not periodic. – Matthew Conroy Dec 26 '12 at 2:12

Here's a try:

For $x$ divisible by 2 but not by 3, put $f(x)=a$.

For $x$ divisible by 3 but not by 2, put $f(x)=b$.

For $x$ divisible by both 2 and 3 (i.e. by 6), put $f(x)=a+b$.

For other $x$ (coprime to 6) put $f(x)=0$.

If we take two coprime integers $m,n$ then suppose e.g. that 2 divides $m$ and 3 divides $n$, then since $m,n$ coprime we have $f(m)=a,f(n)=b$, and also $f(mn)=a+b$ as required by the additive formula. I think the other cases work out also, for example if $m,n$ coprime and $n$ itself coprime to 6 then $f(mn)=f(m)+f(n)=f(m)$ since $f(n)=0$ and multiplying $m$ by $n$ doesn't change whether it is or is not divisible by either of 2,3.

This function $f$ is of period 6, and seems to me to satisfy the additive requirement.

NOTE: In uncookedfalcon's notation this example is $av_2(x)+bv_3(x).$

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