# Nothing in common

Let a and b be distinct integers such that . Show that

### 1. Show that

This is a tricky problem. The key here is that we can re-express the problem via coprime terms to gain some insight.

#### Express as multiples of coprime naturals

Let , then it must be the case that there exists a coprime pair such that . We know that , let’s see what it divides to

(1)

Here, we will employ a natural property of where to show that it must be the case that . In order to claim this, we only need to show that

#### Show that

Using the simple euclidean algorithm, we reduce the problem to

Because by definition , then naturally .

#### Show that

In the same vein, this reduces down to .

#### Show that

Now, because , it must also be the case that . Together with the lemma stated above, this concludes our proof that .

#### To the finish line!

Because , it must be the case that . Then, we can derive the following expression from the problem. Ordering such that so that

This concludes our proof that .

Woosh