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?
 A: If the number $p$ is supposed to be a prime, then your proof is correct. I'd write $0\le 1<p$, 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 $p\mid 1$.
A: Let's go over your proof:


*

*$a=pl\qquad\qquad\;\,$      (OK)

*$a+1=pl+1\quad$ (Makes sense)

*$1\in(0,p)\quad\qquad$ (???)


This last line doesn't make sense to me. Did you somehow mean to derive it from the fist two lines? How does it (in combination with the first two lines) imply that $p\nmid a+1$?
So I would not say that you have a complete proof here. Actually, I the only substantial thing I see is that you rephrase the premiss "$p|a$" to "$a=pl$".

Edit: In the comments the OP said that "$p$ is prime" should have been one of the premisses. This means we can write $1<p$ (I would really prefer this notation over $1\in(0,p)$). 
I would still consider this proof insufficient though. Here is why:
We have the premisses 


*

*$p|a$ (or $a=pl$, whichever you prefer)

*$1<p$


You now claim that this implies that $p\nmid a+1$. But this is the theorem you want to pove. You're basically saying that $A\implies B$, because $A\implies B$. The whole point of proving a theorem is that you make a number of small steps in stead of one big step. 
So if someone, for some reason wasn't convinced that $$p|a\implies p\nmid a+1,\text{ for }1<p$$ then your proof isn't going to convince them. You need to add smaller steps for them. 
The usual way to go is to say:  

Take $a=lp$ and $a+1=kp$. Then $$\underbrace{(a+1)-a=kp-lp=(k-l)p=1.}_\text{small step(s)}$$ This implies that $p|1$, which contradicts that $1<p$. 

Now, instead of saying $A\implies B$, you said $$A\implies \text{(small steps)}\implies B.$$ This is  much more likely to convince someone.
A: 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.... 
