 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Consecutive Primes (Posted on 2005-01-24) Show that the sum of two consecutive odd primes has at least 3 (not necessarily distinct) prime factors. For example,

3+5=2*2*2
5+7=2*2*3
7+11=2*3*3

 See The Solution Submitted by David Shin Rating: 3.1667 (6 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Some thoughts on a particular case Comment 6 of 6 | If we were given the sum of twin primes rather than two consecutive odd primes, then the solution would have been as follows.

Let the twin primes be p and p+2.
For p =3, Sum(3, 5) = 2*2*2, as shown in the problem text.
For p>=5, p must necessarily possess the form 6t-1, where t is a positive integer.
Thus, Sum(p, p+2) = Sum(6t-1, 6t+1) = 12t = 2*2*3*t

Now, for t>=2, t can be either prime or composite.
Thus, for p>=13, Sum(p, p+2) will always contain at least four (not necessarily distinct) prime factors.

Considering that neither (13, 15) nor (15, 17) constitute the set of twin primes, while (17, 19) is a valid set, we can now assert that:

The sum of two twin  primes will always have at least 4 (not necessarily distinct) prime factors, whenever the lesser prime is at least 17.

Edited on May 11, 2007, 1:10 pm
 Posted by K Sengupta on 2007-05-11 13:02:14 Please log in:

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