Consecutive Integers are Coprime

Theorem

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

That is, consecutive integers are always coprime.

Proof 1

$\gcd \left\{{h+1, h}\right\} = \gcd \left\{{h, 1}\right\} = \gcd \left\{{1, 0}\right\} = 1$ from Euclidean Algorithm.

$\blacksquare$

Proof 2

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 $(h + 1)$.

$\blacksquare$