Proof using well ordering principle Assume that the set $I \subseteq\mathbb{Z}$ satisfies the following properties: 
a.) There exists $n \in I$ such that $n \neq 0$
b.) If $m,n \in I$ , then $m+n \in I$
c.) If $m \in I$ and $a \in \mathbb{Z}$, then $am \in I$
This was a hint that was given: Let $J = (m \in I : m >0)$
Prove that there exists $n_0 \in \mathbb{Z}$, such that $I = (kn_0 | k \in \mathbb{Z})$
Seeing as this is from the section of number theory, specifically division algorithm and well ordering principle, I would imagine those are what I am expected to use. 
Intuitively I believe this could be proven by contradiction, but I have no clue how to start and what to do. 
Can someone help?  

Are you familiar with Bezout's Lemma?
If $m,n \in I\subset \mathbb Z$ there are $a,b \in \mathbb Z$ so that $am + bn =\gcd(m,n)$.  Now c; says that $am, bn \in I$ and b; says that $am + bn = \gcd(m,n) \in I$.
Now take the hint:  $J \subset \mathbb N$ and $J$ is not empty (because an $n \ne 0$ is in $I$ and $(-1)n \in I$ so $\pm n \in I$ and one of $n, -n$ must be positive.)  So $J$ has a least element, $n_0$. 
If we take any other element $n \in I$ then $\gcd(n_0, n)\in I$.  But... $0 < \gcd(n_0, n) \le n_0$.  But $n_0$ is the least positive value so $\gcd(n_0, n) < n_0$ is impossible.  So $\gcd(n_0, n) = n_0$.  So all $n\in I$ are multiples of $n_0$ and $I\subset \{kn_0|k\in \mathbb Z\}$.
And by c; $\{kn_0|k\in \mathbb Z\} \subset I$.
So $\{kn_0|k\in \mathbb Z\}=I$.
