Show $p$, $p+1$ and $p+3$ are all primes iff $p=2$
This is, of course, easy to prove one way. That is, assume $p=2$, then $p+1=3$ and $p+3=5$.
However, I am not sure how to prove the reverse? That is, if $p, p+1, p+3$ are all primes, then $p=2$.
How would I be able to show that there isn't a prime greater than 2 that satisfies this condition?
Observe :
Since $p+3$ is prime, it implies that $p$ is not a odd number. (as odd number added with odd number gives you a even number, Simple math!)
hence we conclude from above that $p$ is even. Now as $p,p+1,p+2$ are primes, In particular $p$ is prime.
But $2$ is the only even prime.
Well if $p$ and $p+1$ are both prime, that means that one of them is even, there is only one even prime, namely two. Now if $p+1$ is even, then $p+3$ is also even, however we know that there is only one even prime, so $p+1$ is not even, which leaves $p$ to be the even prime, so $p=2$.
If you wanted to be thorough you could include a simply proof that there is only one even prime.
