Divides
an integer $$n$$ is divisible by another integer $$m$$ if there exists an integer $$k$$ such that $$n = k \cdot m$$. If this is all true we write $$m | n$$.
Divisible iff Integer Quotient
Suppose that $$a \mid b$$ iff $$\frac{b}{a} \in \mathbb{ Z }$$
Suppose that $$a \mid b$$, then there is some $$k \in \mathbb{ Z }$$ such that $$a \cdot k = b$$ therefore $$k = \frac{b}{a}$$, and since $$k \in \mathbb{ Z }$$ then $$\frac{b}{a} \in \mathbb{ Z }$$ as needed.
Not Divisible iff Non-Integer Quotient
$$a \nmid b$$ iff $$\frac{b}{a} \notin \mathbb{ Z }$$
By the fact that $$P \iff Q$$ iff $$\neg P \iff \neg Q$$
1 Divides Everything
Suppose that $$j \in \mathbb{Z}$$, then $$1 | j$$
In the definition of divides take $$k = j$$, then we can see that $$j = j \cdot 1$$ thus we can say that $$1 | j$$
If something Divides One then it is One or Negative One
Suppose $$d \mid 1$$ then $$d = \pm 1$$
We have that $$1 = k d$$ but the only numbers in $$\mathbb{ Z }$$ that have a multiplicative inverse are $$1, -1$$ so $$d$$ must be one of those or else we have a contradiction.
If Zero Divides a Number, then that Number is Zero
Suppose $$0 \mid d$$ then $$d = 0$$
By definition we have $$d = k \cdot 0 = 0$$ so $$d = 0$$
0 is Divisible by Everything
for any $$d \in \mathbb{Z} , d | 0$$
In the definition of divides take $$k = 0$$, then we can see that $$0 = 0 \cdot d$$ thus we can say that $$d | 0$$
Everything Divides Itself
Let $$a \in \mathbb{ Z }$$ then $$a \mid a$$
Clearly $$1 \cdot a = a$$ therefore $$a \mid a$$
Division is Transitive
Suppose $$a \mid b$$ and $$b \mid c$$ then $$a \mid c$$
We have that $$b = k _ a a$$ and $$c = k _ b b$$ therefore $$c = k _ b \left( k _ a a \right)$$ therefore by taking $$k _ c = \left( k _ a k _ b \right)$$ we see that $$a \mid c$$
If you Divide a Number, then You Divide their Multiples
For any $$a, b, c \in \mathbb{ Z }$$ $a \mid b \implies a \mid b c$
Suppose that $$a \mid b$$ therefore there is $$k \in \mathbb{ Z }$$ such that $$b = ka$$, therefore multiply both sides to get $$bc = \left( kc \right) a$$ and therefore $$a \mid bc$$ as needed.
If it Divides it's Less than or Equal with Absolute Values
Suppose that $$n, d \in \mathbb{ Z }$$, such that $$n \neq 0$$ then if $$d$$ divides $$n$$, then $$\lvert d \rvert \le \lvert n \rvert$$

We see that $$n = k d$$ for some $$k \in \mathbb{ Z }$$, since $$n \neq 0$$ then also we must have $$k, d \neq 0$$ specifically note that $$\lvert k \rvert \ge 1$$ so that we have $\lvert n \rvert = \lvert k d \rvert = \lvert k \rvert \cdot \lvert d \rvert \ge 1 \lvert d \rvert = \lvert d \rvert$ so that $$\lvert n \rvert \ge \lvert d \rvert$$

We have to specify $$n \neq 0$$ because we know that everything divides zero, and therefore without that assumption we could use it to claim that since $$-42 \mid 0$$ that $$\lvert -42 \rvert \le 0$$ which is clearly false.

Larger Absolute value Implies no Division
for any $$n, d \in \mathbb{ Z }$$ such that $$n \neq 0$$ if $$\lvert d \rvert \gt \lvert n \rvert$$ then $$d \nmid n$$
This follows by the contrapositive and this
If two Numbers Divide each other then Their Absolute Values are Equal
For any $$a, b \in \mathbb{ Z }$$, if $$a \mid b$$ and $$b \mid a$$ then $$\lvert a \rvert = \lvert b \rvert$$

If $$a = 0$$ then because $$a \mid b$$ then this forces $$b = 0$$ so that $$a = b = 0$$, also if we assumed $$b = 0$$ the same thing occurs.

In the case that neither of $$a, b = 0$$ then recall that if $$a \mid b$$ we have $$\lvert a \rvert \le \lvert b \rvert$$ also since $$b \mid a$$ then also $$\lvert a \rvert \le \lvert b \rvert$$ so that $$\lvert a \rvert = \lvert b \rvert$$.

Divisibility is a Partial Order on the Natural Numbers
Divisibility is a partial order on $$\mathbb{ N } _ 0$$.
It is reflexive because we know that everything divides itself, we know it is transitive, and we also know that it is anti-symmetric, because for any $$n \in \mathbb{ N } _ 0, \lvert n \rvert = n$$ and using this proposition.

Note that we can't quite get a partial order $$\mathbb{ Z }$$ because of the fact that $$-a \mid a$$.

Linearity of Division
Suppose that $$a \mid b _ 1, \ldots , b _ n$$ then $a \mid c _ 1 b _ 1 + \cdots + c _ n b _ n$ where $$a, b _ i, c _ i \in \mathbb{ Z }$$

For the base case of $$n = 2$$ if $$a \mid b _ 1$$ and $$a \mid b _ 2$$ then we know that there are $$k _ 1, k _ 2 \in \mathbb{ Z }$$ such that $$b _ 1 = k _ 1 a$$ and $$b _ 2 = k _ 2 a$$ so that $$c _ 1 b _ 1 + c _ 2 b _ 2 = c _ 1 \left( k _ 1 a \right) + c _ 2 \left( k _ 2 a \right) = a \left( c _ 1 k _ 1 + c _ 2 k _ 2 \right)$$.

For the induction step, assume the statement holds true for $$k \in \mathbb{ N } _ 1$$ and we'll show that it holds true for $$k + 1$$ so suppose that $$a \mid b _ 1, \ldots b _ { k + 1 }$$ by the induction hypothesis we know that $$a \mid c _ 1 b _ 1 + \cdots + c _ k b _ k$$ so we get some $$j ^ \prime \in \mathbb{ Z }$$ such that $$c _ 1 b _ 1 + \cdots + c _ k b _ k = j ^ \prime a$$ and also some $$j ^ { \prime \prime } \in \mathbb{ Z }$$ such that $$b _ { k + 1 } = j ^ { \prime \prime } a$$ and therefore $c _ 1 b _ 1 + \cdots + c _ k b _ k + c _ { k + 1 } b _ { k + 1 } = j ^ \prime a + j ^ { \prime \prime } a = a \left( j ^ \prime + j ^ { \prime \prime } \right)$ so therefore $$a \mid c _ 1 b _ 1 + \cdots + c _ { k + 1 } b _ { k + 1 }$$

Even Number
We say that an integer $$n \in \mathbb{ Z }$$ is even if there is some $$k \in \mathbb{ Z }$$ such that $n = 2k$
A power of an Even Number is Even
If $$n \in \mathbb{ Z }$$ is even, then for any $$m \in \mathbb{ N } _ 1$$ we have $$n ^ m$$ is even
If $$n$$ is even then there is some $$k \in \mathbb{ Z }$$ such that $$n = 2k$$ and therefore $$n ^ m = \left( 2k \right) ^ m = 2 ^ m k ^ m = 2 \left( 2 ^ { m - 1 } k ^ m \right)$$ so that $$n ^ m$$ is also even.
Odd Number
We say that an integer $$n \in \mathbb{ Z }$$ is odd if there is some $$k \in \mathbb{ Z }$$ such that $n = 2k + 1$
A power of an Even Number is Odd
If $$n \in \mathbb{ Z }$$ is odd, then for any $$m \in \mathbb{ N } _ 0$$ we have $$n ^ m$$ is odd

Base case, set $$m = 0$$ then $$n ^ m = 1$$ which is odd. For the induction step assume it holds true for $$k \in \mathbb{ N } _ 0$$ so that $$n ^ k$$ is odd, which means that there is some $$j \in \mathbb{ Z }$$ such that $$n ^ k = 2j + 1$$ therefore $$n ^ { k + 1 } = \left( 2 j + 1 \right) n$$ but $$n$$ is odd so there is some $$j ^ \prime \in \mathbb{ Z }$$ such that $$n = 2 j ^ \prime + 1$$ and thus $n ^ { k + 1 } = \left( 2 j + 1 \right) \left( 2 j ^ \prime + 1 \right) = 4j j ^ \prime + 2j + 2 j ^ \prime + 1 = 2 \left( 2 j j ^ \prime + j + j ^ \prime \right) + 1$ hence $$n ^ { k + 1 }$$ is prime so the statement follows from the principle of induction.

Product of Two Equal Factors means They are Square Roots
Suppose that $$a = bc$$ then $$b = \sqrt{ a }$$ iff $$b = c$$

$$\implies$$ suppose that $$b = \sqrt{ a }$$ then for sake of contradiction if $$b \neq c$$ then if $$c \lt b$$ then we see that $$a = b c \lt b b = \sqrt{ a } \sqrt{ a } = a$$ which is impossible, on the other hand if $$c \gt b = \sqrt{ a }$$ then we see $$a = b c \lt c c \lt \sqrt{ a } \sqrt{ a }$$ which yields the same contradiciton.

$$\impliedby$$ Now suppose that $$b = c$$ therefore $$a = b c = b b$$ therefore $$b ^ 2 = a$$ and so $$b = \sqrt{ a }$$ as needed, moreover in this case note that $$c = \sqrt{ a }$$ as well.

Product Implies Square Root bound on Factors
Suppose that $$a, b, c \in \mathbb{ N } _ 1$$ and that $$b \neq c$$ where $$a = b c$$ then $\left( b \gt \sqrt{ a } \land c \lt \sqrt{ a } \right) \lor \left( b \lt \sqrt{ a } \land c \gt \sqrt{ a } \right)$
Suppose for the sake of contradiction that it was not true, so that we knew that $\left( b \le \sqrt{ a } \lor c \ge \sqrt{ a } \right) \land \left( b \ge \sqrt{ a } \lor c \le \sqrt{ a } \right)$ Suppose that $$b \le \sqrt{ a }$$ and $$b \ge \sqrt{ a }$$ holds true, then $$b = \sqrt{ a }$$ since $$c \neq b$$ then if $$c \lt b$$ we see that $$b c \lt \sqrt{ a } \sqrt{ a } = a$$ which is a contradiction since we know that $$b c = a$$, if it turns out that $$c \le \sqrt{ a }$$ and $$c \ge \sqrt{ a }$$ it is handeled symetrically.

On the other hand if we know that $$b \le \sqrt{ a }$$ and that $$c \le \sqrt{ a }$$, since $$b \neq a$$ then we know that $$b \neq \sqrt{ a }$$, therefore more specifically we see that $$b \lt \sqrt{ a }$$ this implies that $$a = b c \lt \sqrt{ a } \sqrt{ a } = a$$ which is a contradiction. In the case of $$c \ge \sqrt{ a }$$ and $$b \ge \sqrt{ a }$$ it follows in a symmetrical manner.