PART 31. Eleventh and Higher Powers (Update, 1/11/10):
I. Eleventh Powers11.1 Eleven and Twelve terms
(No soln is yet known.)11.2 Fourteen Terms
It is conjectured, as a special
x
for odd 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, x _{1}^{11}+x_{2}^{11}+...+x_{9}^{11 }= y_{1}^{11}+y_{2}^{11}+...+y_{9}^{11}where all terms are positive. 11.5 Twenty Terms
(Update, 1/18/10): Wroblewski
Using a variant of Theorem 5 discussed in
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] [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]
for k = 1,2,3,5,7,9,11 and same conditions 45a II. Twelfth PowersOnly one positive soln is known for the balanced,
x _{1}^{12}+x_{2}^{12}+...+x_{n}^{12 }= y_{1}^{12}+y_{2}^{12}+...+y_{n}^{12}for 6 < n < 12, and it is n = 7. Similar to odd k, it is conjectured that the equal sums of like powers,
x
for even
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)
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 III. Thirteenth PowersNo positive solns are known for the balanced case,
x _{1}^{13}+x_{2}^{13}+...+x_{n}^{13 }= y_{1}^{13}+y_{2}^{13}+...+y_{n}^{13}for 7 < n < 13 though 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) However, Lander (" Three Thirteens", Math. of Comp, Vol 27, Apr 1973) found the multi-grade (k,13,13) for k = 1,3,5,7,9,11,13 as,[1, 9, 25, 51, 75, 79, 103, 107, 129, 131, 157, 159, 173] = [3, 15, 19, 43, 85, 89, 93, 97, 137, 139, 141, 167, 171] which is a result that ought to be improved on by now. IV. Higher PowersUpdate (8/23/09): B. D. Bhargava (bdtara@yahoo.com) gave interesting multi-grade identities
higher than tenth powers. For more details, see section "Maths -- A Unique Equation" of his website: http://bhargavabd.hpage.com. Given any value x, then,A. For k = {1,2,3,...9}:
4(x+2)
^{k} + 2(x+5)^{k} + 2(x+6)^{k} + 4(x+8)^{k} + 9(x+11)^{k} + (x+13)^{k} - 7(x+1)^{k} - 3(x+4)^{k} - 4(x+7)^{k} - (x+9)^{k} - 6(x+10)^{k} - 5(x+12)^{k} = -4(x-2)
^{k} - 2(x-5)^{k} - 2(x-6)^{k} - 4(x-8)^{k} - 9(x-11)^{k} - (x-13)^{k} + 7(x-1)^{k} + 3(x-4)^{k} + 4(x-7)^{k} + (x-9)^{k} + 6(x-10)^{k} + 5(x-12)^{k} - 8x^{k} B. For k = {1,2,3,...10}:
(x-14)
^{k} + 8(x-12)^{k} + 10(x-9)^{k} + 6(x-6)^{k} + 7(x-3)^{k} + 4(x-1)^{k} + 7(x+2)^{k} + 2(x+4)^{k} + 5(x+5)^{k} + 2(x+7)^{k} + 3(x+8)^{k} + 10(x+10)^{k} + (x+11)^{k} + 5(x+13)^{k} = (x+14)
^{k} + 8(x+12)^{k} + 10(x+9)^{k} + 6(x+6)^{k} + 7(x+3)^{k} + 4(x+1)^{k} + 7(x-2)^{k} + 2(x-4)^{k} + 5(x-5)^{k} + 2(x-7)^{k} + 3(x-8)^{k} + 10(x-10)^{k} + (x-11)^{k} + 5(x-13)^{k} C. For k = {1,2,3,...11}:
15(x+2)
^{k} + 5(x+5)^{k} + 6(x+6)^{k} + 13(x+8)^{k} + 29(x+11)^{k} + 7(x+13)^{k} - 26(x+1)^{k} - (x+3)^{k} - 8(x+4)^{k} - 14(x+7)^{k} - 20(x+10)^{k} - 20(x+12)^{k} - (x+14)^{k} = -15(x-2) ^{k} - 5(x-5)^{k} - 6(x-6)^{k} - 13(x-8)^{k} - 29(x-11)^{k} - 7(x-13)^{k} + 26(x-1)^{k} + (x-3)^{k} + 8(x-4)^{k} + 14(x-7)^{k} + 20(x-10)^{k} + 20(x-12)^{k} + (x-14)^{k} - 30x^{k} D. For k = {1,2,3,... 12}:
7(x-14)
^{k} + 22(x-12)^{k} + 33(x-9)^{k} + (x-7)^{k} + 19(x-6)^{k} + 23(x-3)^{k} + 15(x-1)^{k} + 25(x+2)^{k} + 6(x+4)^{k} + 14(x+5)^{k} + 8(x+7)^{k} + 14(x+8)^{k} + 29(x+10)^{k} + 19(x+13)^{k} + (x+15)^{k} =7(x+14) ^{k} + 22(x+12)^{k} + 33(x+9)^{k} + (x+7)^{k} + 19(x+6)^{k} + 23(x+3)^{k} + 15(x+1)^{k} + 25(x-2)^{k} + 6(x-4)^{k} + 14(x-5)^{k} + 8(x-7)^{k} + 14(x-8)^{k} + 29(x-10)^{k} + 19(x-13)^{k} + (x-15)^{k} and so on. One can see the symmetry and how the systems differ when it ends either in an odd or even power. As pointed out by Bhargava, it is possible to construct an
unlimited number of such equations up to any degree k. (End update)--- End ---(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 © 2010
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