Endomorphism from Integers to Multiples

Theorem
Let $\left({\Z, +}\right)$ be the additive group of integers.

Let $\phi: \left({\Z, +}\right) \to \left({\Z, +}\right)$ be a mapping.

Then $\phi$ is a group endomorphism :
 * $\exists k \in \Z: \forall n \in \Z: \phi \left({n}\right) = k n$

Necessary Condition
Let $\phi: \left({\Z, +}\right) \to \left({\Z, +}\right)$ be an endomorphism.

Let $k = \phi \left({1}\right)$.

We have that $n = 1 + \cdots (n) \cdots + 1$ for any positive integer $n$.

Thus:

Also:

Thus:
 * $\forall n \in \Z: \phi \left({n}\right) = k n$

Sufficient Condition
Let $k \in \Z$ such that:
 * $\forall n \in \Z: \phi \left({n}\right) = k n$

Then:

Thus $\phi: \left({\Z, +}\right) \to \left({\Z, +}\right)$ is a group homomorphism from $\Z$ to $\Z$.

Hence by definition $\phi$ is a group endomorphism.