# How can I prove that only there continuous odd prime are $3,5,7$? [duplicate]

How can I prove that the only prime number $p$, such that $p,p+2,p+4$ are primes is 3?

• Hint: All primes except for $2$ and $3$ are of the form $6n\pm1.$ Feb 6, 2016 at 0:15
• @Lucian that's a great hint ... and easy to prove, too.
– John
Feb 6, 2016 at 0:20
• Hint: 1+2 =3. 2+4 =6. Is that too obscure a hint?. Feb 6, 2016 at 3:45
• Even easier. Any prime but 3 is of form 3m +1 or 3m +2. (And 1+2 =3 and 2 +4 =6). Feb 6, 2016 at 3:48

Hint: Assume you have found such a prime, and then look at divisibility by some small, other prime.

• I don't understand... Feb 5, 2016 at 23:05
• It was a hint! The hint was posted one minute ago - you should think about it a little longer than that. To be explicit: There exists a prime $q$ with this property: For any $p$, one of the three numbers $p$, $p+2$ and $p+4$ iis divisible by $q$. Feb 5, 2016 at 23:08

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.

Hint: one of the three numbers $p$, $p+2$, $p+4$ must be divisible by $3$.