Sum of Twin Primes

Date: 09/04/2003 at 22:32:20
From: Alex
Subject: proof for odd prime integers

Given that a and b are two consecutive odd prime integers, prove that 
their sum has three or more prime divisors (not necessarily distinct).

I'm not sure how to represent odd prime numbers in a general statement.

Date: 09/05/2003 at 12:37:51
From: Doctor Paul
Subject: Re: proof for odd prime integers

The first prime must be odd and is hence either one more than a 
multiple of four or is three more than a multiple of four.  

If the first prime is one more than a multiple of four then the second 
prime is three more than a multiple of four.  If the first prime is 
three more than a multiple of four, then the second prime is one 
more than a multiple of four.  

In either case, the sum of the two consecutive odd primes is a
multiple of four.  But notice that this sum cannot be four since the
smallest pair of twin primes are 3 and 5 and 3+5 = 8 which is bigger
than four.

A number that is divisible by four has at least two prime factors -- 
namely two and two.  If the number is bigger than four (which must 
be the case here) then there must be at least one other prime 
factor.  This establishes the desired result.

I hope this helps.  Please write back if you'd like to talk about 
this some more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 09/06/2003 at 11:29:49
From: Alex
Subject: Thank you (proof for odd prime integers)

Thanks Dr. Paul! This really helped! Thanks again!

Date: 01/21/2005 at 06:42:51
From: Federico
Subject: Error in a solution

In http://mathforum.org/library/drmath/view/64087.html it says "if 
the first prime is one more than a multiple of four then the second 
prime is three more than a multiple of four.  If the first prime is 
three more than a multiple of four, then the second prime is one 
more than a multiple of four." 

This is false: for example, 7 and 11 are consecutive primes, but 
both are multiples of four, plus three.

The solution to the problem: if P and Q are consecutive primes, then 
(P+Q)/2 - which is integer - isn't a prime, so (P+Q)/2 = MN, hence 
P+Q = 2MN: three factors, as asked.

Date: 01/21/2005 at 07:15:15
From: Doctor Vogler
Subject: Re: Error in a solution

Hi Federico,

Thanks for writing to Dr Math.  You'll notice that the title of the
page is "sum of twin primes."  Twin primes are pairs of primes whose
difference is two.  That is the question that Dr. Paul answered, and
that is probably what Alex meant, considering that he liked the
answer.  However, when Alex asked the question, he didn't quite make
it clear whether he meant "consecutive odd numbers that are prime"
(i.e. twin primes) or "consecutive prime numbers that are odd" 
(which is how you interpreted it).

But you can still answer the question in the other case.  That is, 
a sum of two consecutive primes (not 2 and 3) is always a product 
of at least three primes.

Of course, one of the prime divisors is 2, since the sum of two odd
numbers is even, so all we have to do is prove that

  (p+q)/2,

the arithmetic mean of consecutive primes p and q, is not prime,
because then it can be factored into at least two primes, and
multiplying by 2 gives a product of at least three primes.

But (p+q)/2 is strictly between p and q, and p and q are 
*consecutive* primes, meaning that there are no primes between them.
So clearly (p+q)/2 cannot be prime.  And that proves it!

If you have any questions about this or need more help, please write
back, and I will try to offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/