Pairs of primes separated by a single number are called prime pairs.
Examples are 17 and 19.
Prove that the number between a prime pair is always divisible by 6 (assuming both numbers in the pair are greater than 6). Now prove that there are no prime triples.

• To prove there are no prime triples...
  Let there be a prime triple, say p, p+2, p+4
  So, all these are prime by assumption.
  
  Exactly one number is divisible by n from n consecutive counting numbers.
  For three consecutive counting numbers exactly one of them is divisible by three.
  
  So, either p or p+1 or p+2 is divisible by 3.
  If p or p+2 is divisible by 3, they are not prime.
  Or if p+1 is divisible be 3, so is (p+1)+3 = p+4.
  
  So, exactly one on p, p+2, p+4 is divisible by 3. Hence atleast on of them is not prime. So, prime triplets cannot exist.
• 15 years agoHelpfull: Yes(6) No(0)
• Proof for number between prime pair is always divisible by 6
  
  Let the prime pairs be p and p+2
  Obviously none of them is divisible by 3, as they are greater than 6 and are prime. Sine exactly one among three consecutive counting numbers have to be divisible by 3, from among p, p+1, and p+2, p+1 must be divisible by 3. Also, p and p+2 are odd. Hence p+2 is even. All even numbers divisible by 3 are also divisible by 6.
  
  So, p+1 is divisible by 6.
• 15 years agoHelpfull: Yes(5) No(0)
• Proof of number between prime pair is divisible by 6: Let pair be p and p+2. Since p and p+2 are not divisible by 3 (as they are prime), p+1 is divisible by 3 (by Pigeonhole principle) and also is even (as p and p+2 being prime are odd). All even numbers divisible by 3 are divisible by 6. Hence proved.
  Proof that there are no prime triplets: This is proved by contradiction. Let there be a prime triple p, p+2 and p+4. By pigeonhole principle, exactly one of p+1 (and hence p+4), p+2 and p+3 (and hence p) are divisible by 3. But this is not possible as p, p+2 and p+4 were assumed to be prime. H
• 15 years agoHelpfull: Yes(0) No(0)