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 ?
I have a feeling that there is something wrong with this question


The "greater than $3$" condition is a bit strange, because the answer is unchanged if $p=3$ is allowed. And if they wanted to exclude negative values of $(p+3)(p-7)$, they should say "greater than 7". But the question as it stands is technically correct. It would perhaps be better like this:


  Which is the greatest natural number that divides $(p+3)(p−7)$ for every odd prime number $p$?


If $p$ is odd, then $p = 2k+1$ for some $k$. So $(p+3)(p-7) = (2k+4)(2k-6) = 4(k+2)(k-3)$, which is divisible by $8$ since one of $k+2, k-3$ must be even. So $8$ divides $(p+3)(p−7)$ for every odd prime number $p$.  

That's half the question. The other half is to show that no natural number $n &gt; 8$ divides $(p+3)(p−7)$ for every odd prime number $p$. Note that if $n$ has this property, then so does every factor of $n$. So it is enough to show that (i) $16$ doesn't have this property; and (ii) no odd prime $q$ has this property.  

To show that $16$ doesn't have this property, just put $p=11$. So suppose now that $q$ is an odd prime. We need to produce another odd prime $p$ that is not equal to $q-3 \pmod q$ and not equal to $7 \pmod q$. Then we can be sure that $q$ doesn't divide $(p+3)(p-7)$. It seems a bit like overkill to me, but we can use Dirichlet's theorem on arithmetic progressions for this: for coprime $a$ and $q$, the arithmetic progression $a + rq$  $(r = 0, 1, 2,...)$ contains infinitely many primes. So we just have to choose $a$ with $1 \le a \le q-1$ that is not equal to $q-3 \pmod q$ and not equal to $7 \pmod q$. If $q=3$, we can choose $a=2$; and if $q \ne 3$, we can choose $a=3$.  

And we're done.
The question is sloppily phrased, because to me it reads more like "Which is the greatest possible natural number that definitely divides $(p+3)(p−7)$, where $p$ is some given prime greater than 3?", in which case the answer is trivially $(p+3)(p-7)$ !

But I imagine (as does everyone else by the look of it) the OP meant "Which is the greatest possible natural number that definitely divides $(p+3)(p−7)$ for every prime $p$ greater than 3?"