### 031: Eleventh and Higher Powers

 PART 31.  Eleventh and Higher Powers    (Update, 1/11/10):     I.  Eleventh Powers   11.1  Eleven and Twelve terms   (No soln is yet known.)  11.2  Fourteen Terms   It is conjectured, as a special balanced case of the Lander-Parkin-Selfridge (LPS) Conjecture, that the equal sums of like powers,   x1k+x2k+...+xnk = y1k+y2k+...+ynk   for odd k where n = (k+3)/2 always has non-trivial solns.  There are parametric identities for all odd k < 11 but, for k = 11, only a single numerical example where all terms are positive is known for (11,7,7).  Interestingly, at least for the first few k, "first" solns have relatively small terms and valid also for k = 1, excepting the one for k = 11,   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 TermsAbout 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 TermsNone is yet known for,x111+x211+...+x911 = y111+y211+...+y911where 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, x112+x212+...+xn12 = y112+y212+...+yn12for 6 < n < 12, and it is n = 7. Similar to odd k, it is conjectured that the equal sums of like powers,   x1k+x2k+...+xnk = y1k+y2k+...+ynk   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. III.  Thirteenth Powers   No positive solns are known for the balanced case, x113+x213+...+xn13 = y113+y213+...+yn13for 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 Powers   Update (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 - 8xk   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 - 30xk   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  ( Original draft: Nov 2006 – Jun 2007 Uploaded: Jun 2009 © 2010     ◄ Previous Page        Next Page ►