1. ## 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?

2. I don't know about (2), but as for (1), let the smaller odd be 2n+1, for some integer n; we see $\displaystyle n\ge{1}$, 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 $\displaystyle n\ge{2}$, it has one or more prime factors, and so the sum has at least three (not necessarily different) prime factors.

--Kevin C.

3. ## 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.

4. I don't know about (2), but as for (1), let the smaller odd be 2n+1, for some integer n; we see , 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 , 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 $\displaystyle p,p'$ can be written as $\displaystyle p=2n+1, p'=2n+3$ for the same $\displaystyle n$.
This implies $\displaystyle p'-p= 2$ for all consecutive $\displaystyle p,p'$

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

5. This is how you do it:

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

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

Conclusion: $\displaystyle p+p'$ has 3 not necessarily different prime-factors.

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

(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.

7. 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

8. Good job guys