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
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
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
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
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!