Demonstrating solutions to two functional equations

Question

Find examples (the more the better) of functions $f: \mathbb{Z} \to \mathbb{C}$ satisfying the relations

1. $f(x+y) = f(x) + f(y)$
2. $f(xy) = f(x)f(y)$

I have only

1. $f(x)= ax$
2. $f(x)= x^a$

This task is from algebra but I don't know why.

Answer 1

You have

$$ f(0+0)=f(0)+f(0) \implies f(0)=0 $$

Also using second condition, for any $y\in \mathbb{Z}$

$$ f(1 \cdot y)=f(1)f(y) $$

Since $\mathbb{C}$ is an integral domain, this can only happen if $f(1)=1$, or $f(y)=0$ for all $y\in \mathbb{Z}$

Notice that $f(n)=f(1+1+\cdots+1)$ where $1$ occurs $n$ times, so what can you say about value of $f(n)$, given above values and condition $f(x+y)=f(x)+f(y)$?

What can you then deduce about $f(-n)$ (note that $f(n-n)=f(0)=0$)

(Your exampple a) fails if $a\neq 0$ or $1$: $f(x)=ax\neq a^2x=f(1)f(x)$)

Answer 2

Answer.

In case a. it is easy to show that $f(n)=nf(1)$. Hence, the $f$ HAS TO BE of the form $f(n)=an$.

In case b. if $\{p_k\}_{k\in\mathbb N}$ are the prime numbers then $$ f(p_1^{a_1}\cdots p_k^{a_k})=f(p_1)^{a_1}\cdots f(p_k)^{a_k}. $$
So if $f\not\equiv 0 $, then it is determined once it is ARBITRARILY defined in the primes and the value $f(-1)$ has to be equal to $\pm 1$.