Problems & Puzzles: Puzzles

Puzzle 96. The least K consecutive odd primes such that the sum of every two of them produces a distinct number

As an easy example, the least set of 4 consecutive odd prime numbers with this property is 3, 5, 7, & 11:

3+5=8
3+7=10
3+11=14
5+7=12
5+11=16
7+11=18

For K=5, the least set is 43, 47, 53, 59 & 61; this set produces 10 distinct sums.

The largest example I have found if a set of K=14 consecutive primes starting with  the prime 334150819 and producing 91 distinct sums

Questions:

1. Get larger sets
2. Do you think that exist at least one set for every K value?

Solutions

Question 1

Jeff Heleen sent the starting primes for K=6 to 13.

Jud McCranie sent the starting primes for K=15 (2,563,210,877) and for K=16 (20,194,423,349).

Question 2

Chris Nash has produced the following extremely beauty and clever existence argument - based in a very accepted conjecture - whose corollary is that for any K value "there exists a set of K consecutive primes such that the sum of every two of them produces a distinct number" .

The existence argument also provides a method to construct those sets no matter that the primes such produced are very large compared to the minimal solution for each K.

This is the delightful argument. Enjoy it!:

"Hola Carlos, An unusual part of a puzzle solution this week, you ask whether or not in Puzzle 96 whether we *think* there are sets of K consecutive primes with all K(K-1)/2 digit sums distinct. Since you said *think* and not * prove*, my answer is yes, provided we assume a certain conjecture is true! The conjecture I will assume is a generalization of Dirichlet's primes in arithmetic progression, and is very similar to the Hardy- Littlewood conjecture. And most people *think* it is true!

Suppose we have k arithmetic progressions with the same common difference, but different starting terms. Unless a simple property holds that would make at least one of these numbers composite, all k of these arithmetic progressions are simultaneously prime infinitely often. The easiest way to avoid this simple property is to choose start terms of the arithmetic progressions that are all prime, and a common difference that is divisible by none of these primes.

Now we can use this result to create arithmetic progressions of *consecutive* primes. Choose any set of prime numbers p_1< p_2.<... p_k, such that p_k<p_1^2. All the numbers between p_1 and p_k that are not in this set are either

1) primes between p_1 and p_k
2) have a factor smaller than p_1

Create a number D which is divisible by all the smallest factors of the numbers between p_1 and p_k. The easiest way to do this is D=p_k#/(p_1.p_2.p_3.....p_k)

Now the numbers Dx+p_1, Dx+p_2..... Dx+p_k are all simultaneously prime for an infinite number of x values. Furthermore, for x>1, all the other numbers between Dx+p_1 and Dx+p_k are all composite (they are of the form Dx+a, where a has a common factor with D). So for an infinity of x values, these are consecutive primes.

Now we will construct a set of k prime numbers such that all their sums are distinct. Choose a prime number p_1 that is greater than 4^(k-1) but less than 4^k. (One must exist, because by Bertrand's postulate, there is always a prime between x and 2x for integers x>1).

By Bertrand's postulate, there is at least one prime number between 2.p_1 and 4.p_1. Call it p_2. It is smaller than 4.p_1. By Bertand's postulate, there is at least one prime number between 2.p_2 and 4.p_2. Call it p_3. It is smaller than (4^2).p_1. ...and so on.... until we have p_k, which is smaller than 4^(k-1).p_1, which is smaller than p_1^2.

Construct the number D as above. Then, if we assume the hypothesis is true, there is at least one value (in fact an infinity of values) of x for which P_1=Dx+p_1, P_2=Dx+p_2, P_3=Dx+p_3..... P_k=Dx+p_k are consecutive primes.

Finally, we prove that these set of consecutive primes all have different pair wise sums. This is very easy to do now. For suppose P_a+P_b=P_c+P_d then by subtracting Dx from every term p_a+p_b=p_c+p_d Suppose p_d is the largest, and suppose p_b>p_a. But p_d>2.p_b>p_a+p_b, and so we have a contradiction.

Therefore, *if we assume the Hardy-Littlewood conjecture is true*, then there exists a set of k consecutive primes such that the sum of every two of them produces a distinct number.

However, notice this is only an *existence* proof! The sequence of consecutive primes it produces are VERY LARGE!

For example, suppose k=4. We need to find 4 primes to start with. The first prime must be between 64 and 256. So we will choose 67. The second prime must be between 2*67 and 4*67, so we will choose 137. The third prime must be between 2*137 and 4*137, so we will choose 277. The fourth prime must be between 2*277 and 4*277, so we will choose 557.

Note they are all primes, and note that the sums of any two of them give six distinct numbers. Now we create the number D=557#/(67.137.277.557). It is a large number.

For any value of x, all the numbers from Dx+67 to Dx+577 are all composite, except possibly for Dx+67, Dx+137, Dx+277, Dx+557. D is not divisible by 67, 137, 277, or 557, so by the Hardy-Littlewood conjecture and its generalization, there are an infinite number of x values for which all four of these numbers are prime. Hence they will be consecutive primes, and all of their sums are distinct.

If you would like an added challenge, try to find a number x such that

Dx+67
Dx+137
Dx+277
Dx+557

are all primes! (I told you they were much larger than the minimum solution, 3, 5, 7, 11!).

[does anybody wants to try?, CR]

Also for fun you may like to predict where this 'existence proof' suggests a solution for K=15. (As a hint to you... the theorem first finds 15 primes, and the largest of these is bigger than 2^42). Note you can modify the construction to use smaller numbers if you wish, but it will still produce very large solutions!

Chris

***

Michael Bell has found and tested that:

5686507*557#/(67*137*277*557)+67
5686507*557#/(67*137*277*557)+137
5686507*557#/(67*137*277*557)+277
5686507*557#/(67*137*277*557)+557

Are all primes, 223 digits each... a kind of large against the least primes with the same property, 3, 5, 7 & 11, as Chris said...

***
Jud McCranie reports the 21/06/2000: "I found no more solutions < 100,000,000,000"

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