Consecutive Integers are Coprime/Proof 2
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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
Let $k \in \Z: k \divides h$.
Also assume $k \divides \paren {h + 1}$.
Thus:
- $\exists a, b \in \N: a k = h, b k = \paren {h + 1}$
Then:
- $\paren {h + 1} - h = b k - a k$
and so:
- $1 = \paren {b - a} k$
Since the integers form an integral domain, $\paren {b - a} \in \Z$.
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 $\paren {h + 1}$.
$\blacksquare$