Assume $p, q$ are prime numbers. Decide whether it is possible for $12p + 12q$ to be equal to$ pq$ . So far I have managed the following
$2(6p + 6q) = pq$
$p$ and $q$ cannot both be even as the only prime number that is even is $2$ and that does not prove this equation.
$p$ and $q$ cannot both be odd as $2$ * anything is even
So that means that one of $p$, $q$ must be odd and the other must be even.
So let $p = 2k$ and $q = 2L+1$
$2(12k + 12L + 6) =  (2k) * (2L + 1) = 24k + 24L + 12 = 4kL + 2k$
Have I done this correctly and do I have sufficient proof to claim that $12p + 12q$ cannot be equal to $pq$ given that both $p$ and $q$ are both prime numbers?
 A: It is obviously false because $pq$ has only two prime factors, but $12(p+q)$ has at least $4$ prime factors.
A: While a good answer has been given, and your own calculations are not wrong, they are just short of the goal, as it isn't apparent why your last result
$$24k+24L+12=4kL+2k$$
is any kind of contradiction. That only becomes clear once you realize that you set $p=2k$, and with $p$ being a prime, $k=1$ must follow. Then your last result becomes
$$24+24L+12=4L+2$$
and this is clearly impossible. One argument is that the left hand side is clearly bigger then the right hand side, another would be that the left hand side is divisible by 4 while the right hand side isn't.
To improve your approach for readability, I suggest just mentioning that $p$ and $q$ can't both be odd first (as you did), this means w.l.o.g $p=2$. Then the equation becomes
$24+12q=2q$, which is impossible due to the "left hand side bigger then right hand side" approach. No need to introduce $k,L$.
A: 0) Let $p,q$ be prime numbers.
1) $p,q \not =2$.
Then $pq$ is odd, while $12(p+q)$ is even, ruled out.
2) Assume $p=q=2$.
Then $pq =4$, ruled out (why?)
3)Assume $p=2$, $q \not =2$.
Then $q= 6(p+q)$, not a prime , ruled out.
Hence?
