For any integer a and any prime number p, if p divides a then p does not divide a+1

Question

I am reading the book Discrete Mathematics by Epp, but I can't understand her proposition 4.7.3 proof (page 210).

The proof claim is: for any integer $a$ and any prime number $p$, if $p$ divides $a$, then $p$ does not divide $a+1$.

Due to divisibility $a = pr$ and $a+1 = ps$ for some integers $r$ and $s$.

It follows that $1 = (a+1) - a$

But this is the part I don't understand. How does Epp come to the step that $(a+1) - a$ equals to $1$?

Answer 1

Perfectly agreeing with Arthur's answer. You're probably overthinking about the step, which is clear as it seems.

You just needed to understand the part where you can just substitute the value of pr and ps for a and a+1, which you did.

The full proof looks something like this, which is proof by contradiction:

We suppose the situation that p divides a and a+1 both Thus we get,

a = pr and a+1 = ps

Now you can just subtract both the equations like this:

(a+1)-a = ps - pr

Using that we get,

1 = p(s-r)

This shows that p divides 1 (as (s-r) is an Integer), which is impossible.

This proves that our assumption is wrong and p can only divide a and not a+1.

Answer 2

I think you're overthinking the wrong step here. $(a+1)-a = 1$ is really obvious. All it says is that the difference between one integer and the next integer is $1$. It's the next step that requires thinking: Substitute $pr$ and $ps$ for $a$ and $a+1$, and you get $$ 1 = (a+1)-a = ps-pr = p(s-r) $$ which means that $p\mid 1$, which is impossible for a prime.