How can I prove that only there continuous odd prime are $3,5,7$? How can I prove that the only prime number $p$, such that $ p,p+2,p+4$ are primes is 3?
 A: Hint: Assume you have found such a prime, and then look at divisibility by some small, other prime. 
A: Out of any three numbers chosen of the form $ p, p+2, p+4$. One of them will be a multiple of three. Here's the reasoning for this.
Out of any three consecutive numbers, one of them has to be a multiple of 3. If none of them were a multiple of 3, then it implies there are two consecutive multiples of three separated by a distance greater than three !
Out of the three numbers, $p-1, p-2, p-3$, one of the three is a multiple of 3.
If p-1 is a multiple of 3, then p+2 is a multiple of 3.
If p-2 is a multiple of three, then p+4 is a multiple of 3.
If p-3 is a multiple of 3, then p is a multiple of 3.
Of all the multiples of 3, only 3 is a prime number. So if we want to check for a sequence of primes of that form, it must include three or else it is composite. We could try the sequence, $1,3,5$. But, 1 is not a prime number. So, that leaves $3,5,7$ as the only sequence.
Hence, proved.
This result is simple, but powerful. However, there is no intuitively obvious reason that there aren't infinitely many primes of the form $p, p+2, p+6$ or $p, p+4, p+6$. Even though the problems looks only slightly harder, it is nearly insolvable ! That's the beauty of mathematics. We don't have to look very far for something we don't know.
A: Hint: one of the three numbers $p$, $p+2$, $p+4$ must be divisible by $3$.
