Problems & Puzzles: Problems

Problem 67. Multigrade Prime Solutions.

Long time ago (1999) we posted the Puzzle 65, related to "Multigrade Relations", name used by Albert H. Beiler, R. K. Guy and others.

A (k,n) multigrade relation is a Diophantine equation system defined as follow:

Σi(aij) = Σi(bij), for i=1 to n and j=1 to k

k is the maximal power used in the Diophantine equations and n is the quantity of terms in each side of the same equations.

Example:

(k=3, n=6): 1+4+5+5+6+9=2+3+3+7+7+8

Which means that:

1+4+5+5+6+9=2+3+3+7+7+8
12+42+52+52+62+92=22+32+32+72+72+82
13+43+53+53+63+93=23+33+33+73+73+83

If it happens that n=k+1 then the "Multigrade Relations" become the "Prouhet-Tarry-Escott problem". Solutions with n=k+1 are said to be "ideal" and are of interest because they are minimal solutions of the problem (Borwein and Ingalls 1994)

Example

(k=3, n=4): 0+4+7+11=1+2+9+10

Which means that:

0+4+7+11=1+2+9+10
02+42+72+112=12+22+92+102
03+43+73+113=13+23+93+103

From now on we will deal only with the "Prouhet-Tarry-Escott problem", that is to say with ideal solutions (n=k+1) such that all the terms in both sides of the Diophantine equations involved are prime numbers.

There is one place in the web's universe devoted to compile this kind of solutions. It is the Chen Shuwen's Equal Sums of Like Powers. And the page concerned with the ideal prime solutions is this one.

There, we may learn that the current state of the art is that there are known solutions only for k= 1 to 9.

In the following table I have extracted some of these solutions (I selected the minimal primes solutions from the results compiled by Shuwen)

k Author [ai] = [bi], i=1 to n, n=k+1
1 Unknown [ 3, 7 ] = [ 5, 5 ]
2 A. H. Beiler, <1964 [ 43, 61, 67 ] = [ 47, 53, 71 ]
3 Carlos Rivera, 1999 [ 59, 137, 163, 241 ] = [ 61, 127, 173, 239 ]
4 Chen Shuwen, 2016 [ 401, 521, 641, 881, 911 ] = [ 431, 461, 701, 821, 941 ]
5 Qiu Min, Wu Qiang, 2016 [ 17, 37, 43, 83, 89, 109 ] = [ 19, 29, 53, 73, 97, 107 ]
6 Qiu Min, Wu Qiang, 2016 [ 83, 191, 197, 383, 419, 557, 569 ] = [ 89, 149, 263, 317, 491, 503, 587 ]
7 Chen Shuwen, 2016 [ 10289, 14699, 27509, 41579, 42839, 65309, 68669, 77699 ] = [ 10709, 13859, 29399, 36749, 46829, 63419, 70139, 77489 ]
8 Chen Shuwen, 2016 [ 3522263, 4441103, 5006543, 7904423, 9388703, 11897843, 13876883, 15361163, 15643883 ] = [ 3698963, 3981683, 5465963, 7445003, 9954143, 11438423, 14336303, 14901743, 15820583 ]
9 Chen Shuwen, 2016 [ 2589701, 2972741, 6579701, 9388661, 9420581, 15740741, 15772661, 18581621, 22188581, 22571621 ] = [ 2749301, 2781221, 6835061, 8399141, 10314341, 14846981, 16762181, 18326261, 22380101, 22412021 ]
10 ??? Not known
11 ??? Not known

I asked Chen Shuwen to suggest a question about the ideal prime solutions. This was his suggestion:

Q. Send your prime solution for j=1 to k, for k=10 and k=11.


On April 8, 2023, Jaroslaw Wroblewski sent the "first known prime solution to the Prouhet–Tarry–Escott problem of degree 11"

[32058169621, 32367046651, 32732083141, 33883352071,
 34585345321, 35680454791, 36915962911, 38011072381,
 38713065631, 39864334561, 40229371051, 40538248081]
 =
[32142408811, 32198568271, 32900561521, 33658714231,
 34978461541, 35315418301, 37280999401, 37617956161,
 38937703471, 39695856181, 40397849431, 40454008891]

All the 24 above terms are (11-digit) prime numbers and the sum of
k-th powers of left terms is equal to the sum of k-th powers of right
terms for any k=1,2,3,...,11.

This solution corresponds to one of the two questions from the Problem 67, from this site, probably posed during the year 2016 or shortly after.

***

On June 17, 2025, Mr. Chen Shuwen sent the following email:

"Dear Rivera

I hope this email finds you well.

I’m pleased to share that my first manuscript is now available online:
https://arxiv.org/abs/2506.11429

In this paper, I have cited some of your outstanding work. If I have inadvertently missed any relevant contributions of yours, please do not hesitate to let me know. Additionally, I’ve included a special acknowledgment of your encouragement and assistance on page 191.

As an independent researcher who has been exploring this beautiful problem for decades, this paper marks my first formal contribution after 40 years of amateur study. I’m certain it has many limitations, and I would be truly grateful for any feedback or suggestions you might offer.

Once again, thank you for your kindness and for your pivotal role in inspiring my work. It remains an honor to have your support"

From the link given above you can get his paper in several formats. In my opinion a great treasury for those interested in this marvelous subject:

A survey of The Prouhet-Tarry-Escott Problem and its Generalizations.

 ***

 

 


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