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:
Σ_{i}(a_{i}^{j})
=
Σ_{i}(b_{i}^{j}),
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^{
}1^{2}+4^{2}+5^{2}+5^{2}+6^{2}+9^{2}=2^{2}+3^{2}+3^{2}+7^{2}+7^{2}+8^{2
}1^{3}+4^{3}+5^{3}+5^{3}+6^{3}+9^{3}=2^{3}+3^{3}+3^{3}+7^{3}+7^{3}+8^{3}
If it
happens that n=k+1 then the "Multigrade Relations" become
the
"ProuhetTarryEscott 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 0^{2}+4^{2}+7^{2}+11^{2}=1^{2}+2^{2}+9^{2}+10^{2} 0^{3}+4^{3}+7^{3}+11^{3}=1^{3}+2^{3}+9^{3}+10^{3}
From now
on we will deal only with
the
"ProuhetTarryEscott 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 (11digit) prime numbers and the sum of
kth powers of left
terms is equal to the sum of kth 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.
***
