Consecutive twin primes

The special thing about all prime numbers greater than or equal to 5 is that they can all be written as 6n + 1 or 6n -1 for some natural number n. This can be proven since 6n - 1 can be written as 6n + 5, and all other natural numbers would be of the form 6n, 6n + 2, 6n + 3, or 6n + 4, which are all composite. Using this we now rewrite the sequence as {6n+1, 6n+3, 6n+5} and also as {6n - 1, 6n + 1, 6n + 3}. But clearly 6n + 3 must be composite! This means that if this sequence of primes were to exist, p would have to be less than 5. So of the primes 2 and 3, we discover that the prime 3 does in fact satisfy that 3 + 2 and 3 + 4 are prime numbers. So we can deduce that the sequence he has given us must be {3, 5, 7}.

$3$ is the only prime divisible by $3$ .

Case 1 : Let's assume that $p$ gives $1$ as remainder when divided by $3$

$p=3m+1$

$p+2=3m+1+2=3(m+1)$

So, $p+2$ is a prime and divisible by $3$

Then, $p+2$ must be $3$

So, $p=1$ but it isn't prime. So, $p \equiv 2,0\pmod{3}$

Case 2 : $p$ gives remainder $2$ when divided by $3$

$p=3n+2$

$p+4=3n+2+4=3(n+2)$

So, $p+4=3$ or $p=-1$

This isn't possible.

So, we must say that $p \equiv 0\pmod{3}$

$p=3x$ and as it is prime,

$p=3$

So we get the numbers as $\boxed{3,5,7}$