Problems & Puzzles: Problems

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 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.

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