Problems & Puzzles: Puzzles

Puzzle 65.- Multigrade Relations

1  + 6  + 8  =  2  + 4  + 9  &
1² + 6² + 8² = 2² + 4² + 9²

is the least example of the so-called multigrade relations.

It's common to express them the following way:

1, 6, 8 = 2, 4, 9; for n=1, 2

Here are others reported:

1, 5, 8, 12 = 2, 3, 10, 11; for n=1, 2, 3
1, 5, 8, 12, 18, 19 = 2, 3, 9, 13, 16, 20; for n=1, 2, 3, 4

There are also published in the literature, multigrade relations involving only prime numbers: 

43, 61, 67 = 47, 53, 71; for n=1, 2
127, 149, 151 = 131, 139, 157; for n=1, 2
281, 281, 1181, 1181 = 101, 641, 821, 1361, for n=1, 2, 3

Results from my own search up today are:

A) Multigrade-prime-relations formed by 6 distinct primes, 3 each side, for n = 1 &2

B) Multigrade-prime-relations with 8 distinct primes, 4 by side, for n = 1, 2 & 3:

Using the known multigrade relation:

1, 79, 105, 183 = 3, 69, 115, 181 = 13, 45, 139, 171, for n = 1, 2 & 3

I have found that adding 58, to each term we get:

59, 137, 163, 241 = 61, 127, 173, 239 = 71, 103, 197, 219, for n = 1, 2 & 3.

C) Multigrade-prime-relations with 8 primes, 4 by side, using 6 consecutive primes, for n = 1, 2 & 3:

Using the known Multigrade relation:

5, 57, 127, 177 = 27, 27, 157, 157, for n = 1, 2 & 3

I have obtained the following one with 6 consecutive primes:

91335911, 91335961, 91336031, 91336081 = 91335931, 91335931, 91336061, 91336061, for  n = 1, 2, 3

D) Multigrade-prime-relations with 10 members, 5 by side, for n=1, 3, 5 & 7

(17, 71, 103, 157, 163 = 31, 47, 121, 143, 169) adding 210 to each term, for n = 1, 3, 5 & 7.

(17, 71, 103, 157, 163 = 31, 47, 121, 143, 169) adding 1830 to each term, for n = 1, 3, 5 & 7.

Would you like to find other / larger / more interesting multigrade-prime-relations?

References:

1.-Recreations in The Theory of Numbers, Albert H. Beiler, p. 163
2.-Concise Mathematics Enciclopedia, Eric Weisstein.
3.-Unsolved Problems in Number Theory, R. K. Guy, p. 143
4.-Equal Sums of Like Powers, Chen Shuwen

Solutions

T.W.A. Baumann found (31/08/99) the following penta-grade-prime-relation:

277, 937, 1069, 2389, 2521, 3181 = 409, 541, 1597, 1861, 2917, 3049
(n = 1,2,3,4,5)

This is his method "I used the (known) integer relation:
1,6,7,17,18,23 = 2,3,11,13,21,22 (n = 0,1,2,3,4,5)
for my solution, multiplying (each term) with 2,4,6,8,... and adding 1,3,5,7,.."

The same way but now using the known hepta-grade relation:
1,5,10,24,28,42,47,51 = 2,3,12,21,31,40,49,50, (n = 1,2,3,4,5,6,7)
he found (1/09/99) the following two hepta-grade-prime-relation:

12251   13841
34511   26561
42461   59951
80621   63131
102881  120371
141041  123551
148991  156941
171251  169661    (n = 1,2,3,4,5,6,7)

84457   88747
114487  105907
191707  221737
268927  226027
311827  354727
389047  359017
466267  474847
496297  492007    (n = 1,2,3,4,5,6,7)

n.b. First column is left side of the relation; second column is the right side of the relation.

***

On Feb 2014 Dr. Jonathan Sondow wrote:

In Puzzle 65.- Multigrade Relations, you say that:

 

1, 6, 8 = 2, 4, 9; for n = 1, 2

is the least example of the multigrade relations.

However, the least example is actually:
 

1, 4, 4 = 2, 2, 5; for n = 1, 2.

A few days later he added:

you say:

‎"Using the known Multigrade relation:

5, 57, 127, 177 = 27, 27, 157, 157, for n = 1, 2 & 3"

‎but the first 5 should be a 7. 

***

On Jan 26, 2021, Adam Stinchcombe wrote:

For consecutive primes I obtain     17, 29, 31, 43=19, 23, 37, 41   for  n=1,2,3.   This is also an improvement for B) Multigrade-prime-relations with 8 distinct primes, 4 by side, for n = 1, 2 & 3, although it does not have three different groups of four primes.

I don’t find consecutive primes through fourth powers up to p=1602912869.

***

On Feb, 10, 2020, Adam wrote again:

I obtain the 10 primes  that form a multi-grade relation up to degree 4:

 

137, 1427, 3347, 3467, 5507=47, 2087, 2207, 4127, 5417  for n=1,2,3,4

 

This was not an exhaustive search so it might not have the smallest total sum.  Given T.W.A Baumann smaller total sum for n=1,2,3,4,5 I suspect a smaller solution can be found for degree 4.

 

With a different programming approach I obtain   11,191,293,593, 647=47,101, 401, 503,683 for n=1,2,3,4.

***

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.