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*