If $p,q$ are prime, $pq\pm 2$, $pq\pm 4$, $pq\pm 6$ cannot be all primes 
Let $p,q$ be distinct primes. Prove that the six integers $pq-2$, $pq+2$, $pq-4$, $pq+4$,$pq-6$, $pq+6$ cannot be all primes.

This is Exercise 5.60 in Chartrand's Mathematical Proofs. The claim is intriguing as for $p=3$ and $q=5$, the four integers $pq-2$, $pq+2$, $pq-4$, $pq+4$ are all primes. I haven't been able to explain why it stops working when $pq-6$ and $pq+6$ are added in the hypothesis.
I think I should look for a proof by contradiction. I don't know where to use the assumption that $p$ and $q$ are distinct primes.
I wonder if there is something peculiar about the prime gaps in this problem. I don't have any other clue.
 A: Let's assume that $pq+2$, $pq+4$ and $pq+6$ are all primes. I'm not considering the other 3 because the same logic will work for them too. One of these three numbers $pq+2$, $pq+3$ and $pq+4$ must be a multiple of 3. But since $pq+2$ and $pq+4$ already are primes therefore $pq+3$ is a multiple of 3. So, $pq+3=3k$ for some natural number $k$.
which in turn implies that $pq+6$ which is just $(pq+3)+3 = 3(k+1)$ is also a multiple of $3$. Which contradicts our assumption that all three of these numbers are primes.
A: If you divide a prime number (other than $2$ or $3$) by 6, the remainder will be $1$ or $5$.  If the remainder were $2$ or $4$ the number would be even, and if you got remainder $3$ the number would be divisible by $3$.
So if $p$ and $q$ are primes greater than $3$, then $pq$ will have remainder $1$ or $5$ when divided by $6$.  If it is $1$ then $pq+2$ cannot be a prime (as it would have remainder 3, when divided by $6$).  If it is $5$ then $pq+4$ cannot be a prime for the same reason.
Thus without loss of generality $p$ must be $2$ or $3$. However if $p=2$, then $pq+2$ will be even and greater than $2$. If $p=3$ then $pq+6$ will be divisible by $3$.
A: Hint
Let $n=pq>6$.
Suppose $3$ does not divide $n-2$, then $n-2 \equiv 1,2 \pmod{3}$, then $n+2 \equiv 2,0 \pmod{3}$. But if $n+2 \equiv 2 \pmod{3}$, then $n+6 \equiv 0 \pmod{3}$.
So if $n-2$ is not divisible by $3$ then one of $n+2$ or $n+6$ has to be divisible by $3$.
Said differently, one of the numbers $n-2, n+2, n+6$ must be divisible by $3$.
Can you take it from here?
