Find all functions $f: \mathbb Z \to \mathbb Z$ such that $f(x+y)=f(x)+f(y)$ Let $x=0\,\, f(x+y)=f(0+y)$ every $y \in \mathbb{Z}$
So we get $f(0)+f(y)=f(y)$ then $f(0)=0$
$0=f(0)=f(x(-x))=f(x)+f(-x)=0$
Let $f(1)=k$ for some $k \in \mathbb{Z}$
$f(n)=f(1+1+1+...+1) (n \text{ many } 1)$
          $=f(1)+f(1)+...+f(1) (n \text{ many } f(1))$
          $=n\cdot k$
Thus $f: \mathbb{Z} \to \mathbb{Z}$ a function such that $f(n)=nk$
I found this but I must also show that there is no other function which satisfies $f(x+y)=f(x)+f(y)$
and I am stuck can somebody help me.
 A: I shall rephrase what I said in my comment, more clearly.
Let $f$ be any function satisfying $f(x) + f(y) = f(x+y)$. Then $f$ must have some value at $1$, so let $f(1)= a$.
Note that $f(k) =f(0+k)= f(0) + f(k)$ for all $k$, hence $f(0)=0$.
Also, note that $f(0) = f(1 + (-1)) = f(1) + f(-1) = a+f(-1)$, so $f(-1) = -a$.
Note that if $k \in \mathbb{Z} > 0$, then $k = 1 + 1 + \ldots + 1$ $k$ times.
Hence, $f(k) = f(1 + 1 + \ldots 1) = f(1) + f(1) + \ldots + f(1) = kf(1)=ak$.
Similarly, $f(-k)= kf(-1) = -ak = (-k)a$.
Hence, the value of such a function at $1$ fixes the value of the function at every other integer.
Now, it is easy to see that $f(k) = ak$ for every $k \in \mathbb Z$, since we proved it above for $k>0$, and for all negative integers $-k<0$ (and it is anyway true for $k=0$). Hence, there can be no function different from this, because I showed above that every function satisfying this property must be of the form above. I did not assume anything about the function other than $f(x+y) = f(x) + f(y)$ at all, so that means before solving the question, that $f$ could have been any random function, different from $f(k) = ak$ also! As I have showed, that is not the case.
Hence, any $f$ satisfying the constraints is of the form $f(k) = ak$, where $a = f(1)$.

More precisely, note that with $\mathbb Z$ as a group under addition,$f$ satisfies nothing but what homomorphism must between $\mathbb Z$ and $\mathbb Z$ . Hence, this also describes all homomorphisms from $\mathbb Z$ to itself. 
If we were working in such a setting, then indeed it would suffice to say that $\mathbb Z  = \langle 1\rangle$, so every homomorphism of $\mathbb Z$ to itself is determined by where $1$ goes.
