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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.

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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.

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  • $\begingroup$ Is the correct process to substitute $ 2n+1 $? I have done this with the first case, $ (2(2n+1) - (2 + (2n+1)) $ and resulted in $ 2n - 1 $ and as a result and integer $ n $ would result in an odd number. $\endgroup$ – Bayyls Mar 29 '17 at 8:10
  • $\begingroup$ Yep. Ant the second gives you $(2k +1)(2j+1) - ((2k+1) - (2j+1))$. But once you proven odd+odd = even and odd*odd = odd etc. once, you can refer to them again $2p$ is even. So $2p - 2$ is even so $2p - 2 - p$ is odd. $pq$ is odd. so $pq - p$ is even. So $pq-p-q$ is odd. $\endgroup$ – fleablood Mar 29 '17 at 8:19
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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.

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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 $2$.

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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.

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  • $\begingroup$ "Generally this is not true if you allow the primes to be the same." Um, right... if you have two distinct things that are the same.... $\endgroup$ – fleablood Mar 29 '17 at 8:06
  • $\begingroup$ @fleablood Yes, that was basically what I meant (that it doesn't hold fully in the more general sense - the more general is in the last section). Perhaps I could have phrased it differently? $\endgroup$ – skyking Mar 29 '17 at 10:28

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