Which is the greatest possible natural number that divides $(p+3)(p-7)$, where $p$ is a prime number greater than $3$? 
Which is the greatest possible natural number that definitely divides 
  $(p+3)(p-7)$, where $p$ is a prime number greater than $3$?

This one is from my module, comes as a fill in the blanks with no answer. I have a feeling that there is something wrong with this question, since for $p=5$, one has $(p+3)(p-7)=-16$.

ADDED:
As mentioned, in one answer I tried substituting $p=5,7,9,11,13,17,19,...$ and spotted that the greatest number is $8$. I am just wondering .. is there any other way (probably using modulus) to directly find this number without actually substituting ?

If $p$ is prime it can be written $p=4a+1$ or $p=4a+3.$  
Substituting the first of these gives
$$(p+3)(p-7)=p^2-4p-21 = 16a^2+8a+1-16a-4-21 = 16a^2-8a-24 = 8(2a^2-a-3)$$
and the other gives
$$(p+3)(p-7)=p^2-4p-21 = 16a^2+24a+9-16a-12-21 = 16a^2+8a-24 = 8(2a^2+a-3)$$
and so $(p+3)(p-7)$ is divisible by $8$ either way.  
Edit
To show that 8 is the largest such divisor, we must show that $2a^2-a-3$ is not divisible by the same prime for all values of $a$. Setting $a=4\text{ and }5$ produces 25 and 42 respectively which have no prime factors in common.  
And we must also show that $2a^2+a-3$ is not divisible by the same prime for all values of $a$. Setting $a=4\text{ and }5$ again produces 33 and 52 respectively which also have no prime factors in common.  
