Sum of Twin PrimesDate: 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/ |
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