There is a list of distinct numbers. There are at least two numbers in the list. Each number in the list is a sum of two primes where the difference between the two primes is 2. What is the minimum possible greatest integer that divides all the numbers in the list? What can you say about the maximum possible one?

Sunday, September 18, 2011

Please email your solutions to puzzles@cotpi.com.

### 2 comments

### Credit

This is an original puzzle from cotpi.

### Further reading

The following is a list of resources on related topics:

- Sebah, Pascal; Gourdon, Xavier "Introduction to twin primes and Brun's constant." Mathematical Constants and computation. 30 July 2002. 21 Sep. 2011 <http://numbers.computation.free.fr/Constants/Primes/twin.pdf>.
- Weisstein, Eric W. "Twin Primes." Wolfram MathWorld. 21 Sep. 2011. Wolfram Research. 21 Sep. 2011 <http://mathworld.wolfram.com/TwinPrimes.html>.
- Weisstein, Eric W. "Twin Prime Conjecture." Wolfram MathWorld. 21 Sep. 2011. Wolfram Research. 21 Sep. 2011 <http://mathworld.wolfram.com/TwinPrimeConjecture.html>.

## Karthik Krishnamurthy solved this puzzle:

Minimum possible GCD: 4.

Every twin prime pair greater than (3, 5) can be written as (6k − 1, 6k + 1) where k is a positive integer. Sum of twin primes is therefore of the form: 12k for some integer k ≥ 1. Thus, the list formed by adding twin primes other than (3, 5) is {12N

_{1}, 12N_{2}, …, 12N_{n}} where n is a positive integer and integer N_{i}≥ 1 (1 ≤ i ≤ n).Since, the smallest number possible in such a list is 5 + 7 = 12, the minimum possible GCD for such a list of numbers is 12.

Can we get a list whose GCD is smaller than 12? Yes. Include 3 + 5 = 8. The list must be of the form {8, 12N

_{1}, 12N_{2}, … 12N_{n}}. The GCD of the numbers in this list is 4. Since the smallest twin prime pair is (3, 5) we cannot make a list of sum of twin primes with a GCD smaller than 4.The maximum possible GCD must be a multiple of 12 which is obvious from the fact that the sum of twin primes are of the form 12k where k is a positive integer. If there are infinitely many twin primes (twin prime conjecture), it seems to indicate one might always find a list with sum of twin primes which have a GCD that is higher than a given list of sum of twin primes but this is just a guess.

## Ted Schuerzinger solved this puzzle:

For the first question, the answer is 4. One of the twin prime pairs is 3 and 5.

If the pairs don't include 3 and 5, then the lower prime n ≡ 2 (mod 3), and the higher prime n + 2 ≡ 1 mod 3. These two sum to 3 ≡ 0 (mod 3). The primes, both being odd, have to be 1 (mod 4) and 3 (mod 4), which sums to 4 ≡ 0 (mod 4). A number divisible by both 3 and 4 is obviously divisible by 12.

You can say almost nothing else about the maximum possible integer until the question over whether or not there is an infinite number of twin prime pairs is answered. That's a bit beyond me. I don't even have a lovely proof that's just a bit too big to fit in the margin of this post.