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We have to prove that if the difference between two prime numbers greater than two is another prime,the prime is $2$. It can be proved in the following way.
1)$Odd -odd =even$.
Therefore the difference will always even.
2)The only even prime number is $2$.Therefore the difference will be $2$ if the difference between primes is another prime.
I am looking for more proofs to this theorem.Any help will be appreciated.
The proof you provided is fine way of proving the proposition. Here is an alternate proof, set up as a contradiction,
Suppose that the difference between two odd primes $a=2n+1$ and $b=2m+1$ is an odd prime, $c$.
$$a-b=c \implies (2n+1)-(2m+1)=c \implies 2(n+m)=c$$
Therefore, $2|(a-b)$, a contradiction.
It is therefore impossible for the difference of two odd primes to be an odd. This means that the difference of two odd primes must be even. The only even prime is 2.
So, if the difference of two odd primes is a prime, then it must be two.
Well, since the first prime must be odd, then:
(1) If the "other prime" is odd their difference is even and the only even prime is two
(2) If the "other prime" is even then it is two and we have twin primes, like $\,13-2=11\,$
Here is a proof by contradiction. Let $p, q, r$ be primes greater than $2$ with $p+q=r$
$p$ and $q$ are odd, so $p=2a+1, q=2b+1$ and $$r=p+q=2a+1+2b+1=2(a+b+1)$$
But this is a factorisation of $r$ and contradicts the fact that $r$ was chosen to be prime.
