Consecutive Integers are Coprime

Theorem
$$\forall h \in \Z$$, $$h$$ and $$h + 1$$ have only two common factors, $$1$$ and $$-1$$.

Proof
Let $$k \in \Z: \; k \backslash h$$.

Also assume $$k \backslash (h + 1)$$.

Thus, $$\exists a, b \in \N: a \cdot k = h, \; b \cdot k = (h + 1)$$

Then $$(h + 1) - h = b \cdot k - a \cdot k$$.

$$1 = (b - a) \cdot k$$.

Since the integers form an integral domain, $$(b - a) \in \Z$$.

Thus either $$k = 1$$ and $$b - a = 1$$, or $$k = -1$$ and $$b - a = -1$$.

Therefore, only $$1$$ and $$-1$$ can be factors of both $$h$$ and $$(h + 1)$$.