How many maps $\phi:\Bbb{N}\to\Bbb{N}$ are there with $\phi(ab) = \phi(a)+\phi(b)$?

I am stuck on the following problem when I was trying to solve an entrance exam paper:

How many maps $\phi \colon \Bbb N \cup \{0\} \to \Bbb N \cup \{0\}$ are there with the property that $\phi(ab)=\phi(a)+\phi(b)$ for all $a,b \in \Bbb N \cup \{0\}$ ?

The options are as follows :

1. none

2. finitely many

3. countably many

4. uncountably many

Option 1 is not possible as if we take $a=1,b=0$ then we get $\phi(1)=0$ which is possible. But, I am not sure about the other options and could not decide which one holds good.

Can someone explain?

Solution 1:

I may be missing something- but it seems that the only map with this property is identically $0$. The proof is as follows:

$\phi(0.0)=\phi(0)+\phi(0)$ which implies that $\phi(0)=0$.

Now, for any $a \in \mathbb{N}$, $\phi(a.0)=\phi(a)+\phi(0)$ which means $\phi(0)=\phi(a)+\phi(0)$ which implies $\phi(a)=0$.

Solution 2:

Certainly the map $\phi(n) = 0$ works.

Taking $a=0$ gives $\phi(0) = \phi(0) + \phi(b)$, whence $\phi(b) = 0$. So in fact the identically zero map is the only solution.

Solution 3:

Hint: Consider $\phi(0\cdot b)$.