Using proof by contradiction for problem

Question

The question is: Show using proof by contradiction that there are no two primes p1 and p2 greater than 2 such that p1 − p2 is odd. Remember that a number k is prime if and only if its only factors are k and 1 (and 1 is not prime).

I have no idea how to start this and would appreciate any help.

Answer 1

Let’s first prove a lemma (a little result used to prove a bigger theorem).

All prime numbers greater than 2 are odd.

Proof:

Suppose a prime number p great than 2 is not odd. Then p is even. So p=2k, for some natural number k. But k>1, since p>2. Thus, k>1 is a factor of p. Thus, p has at least one other factor besides p and 1. Thus p is not prime, as assumed. Thus, we have a contradiction. So, all prime numbers greater than 2 are odd.

Proof by contradiction:

Now suppose p and q are two prime numbers, both greater than 2 such that p−q is odd. Since p and q are prime numbers greater than 2, they must be odd, by the above lemma. Thus, there exist integers m and n such that p=2m+1 and q=2n+1. Now, p - q = (2m+1) - (2n+1) = 2(m - n), which is even. This contradicts the assumption that p-q is odd. This completes the proof by contradiction.

Answer 2

Any prime greater than 2 is odd. So p_{1}, p_{2} are odd numbers. But it is trivial to check that difference of odd numbers is always even. This completes the proof.