031: Eleventh and Higher Powers

k = 1,3: [1, 5, 5] = [2, 3, 6]

k = 1,5; [5, 6, 6, 8] = [4, 7, 7, 7]; (Subba Rao, 1934)

k = 1,7; [2, 12, 15, 17, 18] = [8, 8, 13, 16, 19]; (LPS, 1967)

k = 1,9; [1, 13, 13, 14, 18, 23] = [5, 9, 10, 15, 21, 22]; (LPS, 1967)

k = 0,11; [46, 52, 115, 119, 249, 566, 614] = [127, 152, 175, 212, 441, 487, 629]; (Nuutti Kuosa, 2004)

As k = 11 was not found by an exhaustive search (?), it may be possible there is a smaller soln valid for both k = {1, 11}, just like the rest in this list. In fact, the one for k = 9 belongs to a family found by Bremner and Delorme (2009) that is for k = 1,3,9. No example is known for (13,8,8) but, if indeed first solns are relatively small, it may be now feasible to find it. The closest is a (13,9,9) with a zero term,

k = 1,13: [0, 18, 24, 40, 45, 58, 64, 64, 72] = [15, 23, 29, 32, 43, 52, 53, 65, 73]; (Greg Childers, 2000)

11.3 Sixteen Terms

About 47 positive solns are known to the balanced case and several, perhaps not surprisingly, are good for k = 1,11,

k = 1,11; [5, 8, 20, 30, 30, 62, 63, 72] = [7, 13, 15, 23, 37, 56, 69, 70]; (Nuutti Kuosa, 2002)

11.4 Eighteen Terms

None is yet known for,

x₁¹¹+x₂¹¹+...+x₉¹¹ = y₁¹¹+y₂¹¹+...+y₉¹¹

where all terms are positive.

11.5 Twenty Terms

(Update, 1/18/10): Wroblewski

Using a variant of Theorem 5 discussed in Equal Sums of Like Powers, he used the 10th power [k,6,6] identity to get an 11th power [k,10,10] using,

NewLeftTerms = {OldLeftTerms+c, OldRightTerms-c}

NewRightTerms = {OldLeftTerms-c, OldRightTerms+c}

Normally, this doubles the number of terms but since four terms cancel out, one gets,

[a+3b+2c, -a+3b-2c, 2a-8b-c, -2a-8b+c, -8a-2b-c, -8a+2b+c, 3a+b-c-d, 3a-b+c-d, 3a+b-c+d, 3a-b+c+d]^k =

[a+3b-2c, -a+3b+2c, 2a-8b+c, -2a-8b-c, -8a-2b+c, -8a+2b-c, 3a+b+c-d, 3a-b-c-d, 3a+b+c+d, 3a-b-c+d]^k

for k = 1,2,3,5,7,9,11 and same conditions 45a²-11b² = c², 45b²-11a² = d². Naturally, c has just been negated in the RHS. The process can be continued, with c replaced by d in the transformation, to find a [k,16,16] for k = 1,2,3,4,6,8,10,12, since eight terms cancel. (End update.)

II. Twelfth Powers

Only one positive soln is known for the balanced,

x₁^k+x₂^k+...+x_n^k = y₁^k+y₂^k+...+y_n^k

for even k where n = (k+2)/2 always has non-trivial solns. Identities are known for all even k < 12, but there is only one numerical example for (12,7,7). "First" solns have relatively small terms:

k = 2: [1, 7] = [5, 5]

k = 4: [2, 4, 7] = [3, 6, 6]

k = 6: [2, 2, 9, 9] = [3, 5, 6, 10]; (Subba Rao, 1934)

k = 8: [1, 10, 11, 20, 43] = [5, 28, 32, 35, 41]; (LPS, 1967)

k = 10: [5, 23, 34, 34, 85, 92] = [16, 25, 28, 32, 71, 95]; (Randy Ekl, 1997)

k = 12: [3, 37, 42, 48, 88, 89, 95] = [30, 54, 73, 73, 74, 77, 99]; (Greg Childers, 2000)

No example is known for k = 14 as a (14,8,8) though, if the pattern continues, the first soln might have terms just in the lower end of 3 digits. However, there is a [14,9,9],

k = 14: [8, 14, 20, 25, 47, 83, 110, 113, 115] = [23, 27, 38, 42, 69, 73, 92, 104, 121]; (Jaroslaw Wroblewski, 2003)

which in fact is good for k = 2,14.

(Postscript: I hope you enjoyed this book and that it will motivate some of you to giving at least a little time to some of the problems I raised here. In particular, I really wish someone can give another identity for the multi-grade octic case k = 2,4,6,8, or crack the ninth degree one for k = 1,3,5,7,9. Then, of course, there are the higher powers. I really want to know if it is doable. I tried, but couldn't, but there must be someone out there who might be able to so. Anyway, feel free to send remarks to the email address given below.)

Tito Piezas III (tpiezas@gmail.com)

Original draft: Nov 2006 – Jun 2007

Uploaded: Jun 2009

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