# two consecutive primes

• January 26th 2010, 06:09 AM
melissaanderson
two consecutive primes
Congruence cases.

(1) Let b be the sum of two consecutive odd numbers that are primes (for example, b = 11 + 13). Prove that b has at least three prime factors (not necessarily different).

(2) Let b be the sum of two consecutive primes p and p', where p' > p >= 3 (for example, b = 89 + 97). Prove that b has at least three prime factors (not necessarily different).

Up for the challenge?
• January 26th 2010, 07:32 AM
TwistedOne151
I don't know about (2), but as for (1), let the smaller odd be 2n+1, for some integer n; we see $n\ge{1}$, and the other odd number is 2n+3. Thus, their sum is (2n+1)+(2n+3)=4n+4=4(n+1)=2*2*(n+1);
as $n\ge{2}$, it has one or more prime factors, and so the sum has at least three (not necessarily different) prime factors.

--Kevin C.
• January 26th 2010, 07:45 AM
Manx
one of two
Let's consider #1.
Consecutive odd integers that are primes implies consecutive primes, neither of which is 2. Hence, both primes are odd.
Therefore, b is even. It has a factor of 2.
b=p1 + p2 = 2k.
Then k = (p1+p2)/2, the average of p1 and p2.
Assuming p1<p2, k = p1/2 + p2/2 > p2/2 + p2/2 = p2.
Similarly, k > p1.
But if p1 < k < p2, then k is the even integer that lies between the two odd primes. Therefore k = 2n for some integer, n. n > 1 since n=1 implies k=2 implies p1=1 and p2=3 which is a contradiction.

I think you can take a similar approach in #2.
Good luck.
• January 26th 2010, 09:49 AM
Dinkydoe
Quote:

I don't know about (2), but as for (1), let the smaller odd be 2n+1, for some integer n; we see http://www.mathhelpforum.com/math-he...e4f02030-1.gif, and the other odd number is 2n+3. Thus, their sum is (2n+1)+(2n+3)=4n+4=4(n+1)=2*2*(n+1);
as http://www.mathhelpforum.com/math-he...0368e1f9-1.gif, it has one or more prime factors, and so the sum has at least three (not necessarily different) prime factors.

--Kevin C.
You're assuming that 2 different consecutive primes $p,p'$ can be written as $p=2n+1, p'=2n+3$ for the same $n$.
This implies $p'-p= 2$ for all consecutive $p,p'$

(p=13, p'= 17 are also consecutive primes!)
• January 26th 2010, 10:36 AM
Dinkydoe
This is how you do it:

Observe that for any 2 consecutive primes $p,p'$, we have $2|p+p'$ and

$p<\frac{p+p'}{2} < p'$ hence $\frac{p+p'}{2}$ is composite since $p,p'$ were consecutive primes.

Conclusion: $p+p'$ has 3 not necessarily different prime-factors.
• January 26th 2010, 02:20 PM
DKman
Quote:

Originally Posted by Dinkydoe
You're assuming that 2 different consecutive primes $p,p'$ can be written as $p=2n+1, p'=2n+3$ for the same $n$.
This implies $p'-p= 2$ for all consecutive $p,p'$

(p=13, p'= 17 are also consecutive primes!)

The first question asks for two consecutive odd numbers that are prime. Not two consecutive prime numbers that are odd. So I believe he is looking at the problem correctly.

Your solution does apply to both cases though.
• January 26th 2010, 02:58 PM
Dinkydoe
Quote:

The first question asks for two consecutive odd numbers that are prime. Not two consecutive prime numbers that are odd. So I believe he is looking at the problem correctly
.

Ai, you're right, I'm sorry Kevin C. I misinterpreted the question (Giggle)
• January 27th 2010, 11:10 AM
melissaanderson
Good job guys