Proof: The difference between the product of two distinct prime numbers and their sum must be odd. Prove: The difference between the product of two distinct prime numbers and their sum must be odd.
I attempted to disprove the hypothesis by finding two distinct primes: 
$i, k$ where $(i \cdot k ) - (i - k) \equiv 2n $ 
As I have not been able to find a pair of primes to satisfy my equality, should I instead be trying to prove with a contrapositive or contradiction? If so how would go about changing the original statement.
 A: The only important thing about primes is that primes are, with the exception of $2$, odd.
So either you have $2$ and an odd prime $p$ and need to show $2p - (2 +p)$ is odd.
Or you have $p$ and $q$ two odd prime and you need to show $pq - (p + q)$ is odd.
...
Actually, if you have any two numbers, $a$ and $b$ and they are not both even (at least one is odd) then $ab - (a+b)$ is odd.
A: I might do an even/odd analysis, breaking it into cases:
Case 1: i is 2.
Case 2: k is 2.
Case 3: Neither are 2. 
A: Check out the cases where $p=2$. Now if $p,q$ are two distinct prime numbers, then $p\equiv 1 $ modulo 2 and $q\equiv 1$ modulo 2. Then, reduce $pq-p-q$ modulo $2$.
A: Generally this is not true if you allow the primes to be the same. If the primes are $2$ you have that $2\cdot 2=4$ and their sum is $2+2=4$ the difference is then $0$ which is even. In fact we have if $p$ and $q$ are primes and not both are $2$ we have that $pq-(p+q)$ is odd.
Now we see that if neither of the primes is $2$ they are both odd and you have that the product is odd and the sum is even.
If OTOH one is $2$ and the other is $p\ne 2$ you have that the product is $2p$ and the sum is $p+2$ the difference is $2p-(p+2) = p-2$, but since $p\ne 2$ we have that $p$ is odd and therefore $p-2$ too.
