If $p | a$, prove that $p \nmid (a+1)$ Suppose that $p | a$.  Prove that $p \nmid a + 1$.

$a = p l$

$a + 1 = (pl) + 1$

$1 \in (0, p)$

As such, $p \nmid a + 1$ by definition of divisibility.

Does this look correct?
If the number $p$ is supposed to be a prime, then your proof is correct. I'd write $0\le 1&lt;p$, rather than $1\in(0,p)$.

Your proof appeals to uniqueness of the remainder, rather than definition of divisibility, and you should mention it.



A different proof is also easy: if $a=pl$ and $a+1=pk$, then
$$
1=(a+1)-a=pk-pl=p(k-l)
$$
which implies $p\mid 1$.
I shall try it by contradiction.... 

Let p|a and p|a+1
Let q be a prime divisor of p....then q also divides a and a+1.... 
Hence it divides ua + v(a+1); for u and v integers..... 
In particular q|(-a+a+1)....
i.e. q|1
But q was a prime..... Hence q >=2.....
Hence q cannot divide 1(in integers) ....
A contradiction...... Hence p can not divide any two consecutive integers until and unless p =1.....

U can also try it using congruency....