Show that the sum of two consecutive odd primes has at least 3 (not necessarily distinct) prime factors. For example,
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