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.

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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. 

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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.

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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.

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