029: Ninth Powers

PART 12.  Ninth Powers 


9.1  Nine terms

9.2  Ten terms

9.3  Twelve terms

9.4  Fourteen terms
9.5  Sixteen terms
9.1  Nine Terms
It is conjectured as a special case of the Lander-Parkin-Selfridge Conjecture that, for odd k > 3, the eqn,

x1k + x2k + …+ xkk = 0

has non-trivial solns in positive and negative integers.  Three are known for k = 5, but none have yet been found for k = 7,9, or higher.


9.2  Ten Terms
Results for 9th powers and higher with a minimal number of terms came only in the last decadeFor an equal sum of five 9th powers,


x1k+x2k+x3k+x4k+x5k = y1k+y2k+y3k+y4k+y5k   (eq.1)


the first for k = 9 was found by Ekl in 1997 and is given by,


{192, 91, 101, 26, 30} = {12, 180, 17, 175, 116}


Surprisingly this also satisfies a lot of side conditions, namely x1 = x2+x3 = y1+y2 = y3+y4.  The first for k = 1,3,9 was found by Wroblewski in 2004,


{51, 253, 412, 600, 624} = {100, 187, 429, 603, 621}


which also satisfies x1+x2+x3 = y1+y2+y3; x4+x5 = y4+y5.  While an analogous k = 1,3,5 is trivial, there is a non-trivial k = 1,3,7 and now a k = 1,3,9, so it is tempting to speculate there is a k = 1,3,n for all odd n > 5.  There are now 22 known solns to ten 9th powers equal to zero, though Wroblewski's is the only one that is multi-grade, or even valid for k = 1.  See his tables here.
a)  Whether there are nine 9th powers equal to zero remains unknown.
b)  There is yet no solution to x1+x2+x3+x4+x5 = y1+y2+y3+y4+y5 = 0, unlike for 5th and 7th powers where analogous systems are known.
c)  In general, whether there are k positive kth powers equal to a kth power for k > 8 is also unknown, though Wroblewski has come close with ten 9th powers and twelve 10th powers:
429 + 999 + 1799 + 4759 + 5429 + 5749 + 6259 + 6689 + 8229 + 8519 = 9179
6210 +11510 +17210 +24510 +29510 +53310 +68910 +92710 +101110 +123410 +160310 +168410 = 177210
(Update, 2/18/10):  On a hunch, this author re-checked Wroblewski’s soln for (k.5.5) for k = 1,3,9 to see if there was a partition such that it would be good for k = 2 as well.  It turns out there was.  Thus,


[51, 253, 412, -621, 600]k = [187, 100, 429, 603, -624]k,  for k = 1,2,3,9


with terms obeying the constraints,


1. x1+x2+x3 = y1+y2+y3

2. x4+x5 = y4+y5

3. (x1-y1)(x2-y2) = -(x3-y3)(x4-y4)

4. (x1-y1)(x2-y2) = (x3-y3)(x5-y5)


with the last two suggested by Wroblewski, though [4] is just a consequence of [2] and [3].  Why this “numerical curiosity” obeys such symmetric constraints is unknown.  (Note that the difference xi-yi for all five pairs is divisible by 17.)  (End update.)

9.3  Twelve Terms
(Update, 1/11/10)For twelve terms, or an equal sum of six 9th powers,


x1k+x2k+x3k+x4k+x5k+x6k = y1k+y2k+y3k+y4k+y5k+y6k


the multi-grade system k = 1,3,5,7,9 used to have just one known soln, but Wroblewski increased the total to five in 2009.  (If the special case x1 = 0 is found, this will lead to an ideal soln of deg 10 of the Prouhet-Tarry-Escott problem.)  Interestingly, all but one have a re-arrangement such that it is good for k = 2 as well, 


[-269, -173, -7, 29, 311, 313] = [-247, -193, -59, 91, 289, 323]
[57, 679, 1293, -115, -925, -1279] = [299, 767, 1205, -399, -995, -1167]
[407, 163, 341, -37, -371, -119] = [221, 311, 403, -23, -181, -347]  
[365, 1115, 1325, -305, -731, -1037] = [13, 23, 1319, 1177, -689, -1111]
[43, 161, 217, 335, 391, 463] = [85, 91, 283, 287, 403, 461]   (only for k = 1,3,5,7,9)
The first was found by Shuwen in 2000 after two months of computer time!  Wroblewski derived the second one from a special k = 2,4,6,8 eqn, and found the next three via a numerical search.  (End update
Two things can be noted.  First, analogous to the situation for even systems k = 2,4,…2n which, if terms are signed, can be made valid for k = 1, it seems odd systems k = 1,3,…2n+1 can be made true for k = 2 since one can “move” terms around and there might be a balanced partition,


x1k+x2k+…+xmk = y1k+y2k+…+ymk


also true for k = 2.  The consequence of letting a system valid for an even power is that one can no longer transpose them arbitrarily and it may "behave" more predictably.  Recall that we extended a result for fifth powers and gave a conjecture for the seventh as,


"Let Fk:= x1k+x2k+x3k+x4k+x5k – (y1k+y2k+y3k+y4k+y5k).  

If Fk = 0 for k = 1,2,3,5,7, then (x12+x22+x32+x42+x52)(11F9) = 9F11."


Its analogue for ninth powers is,


"Let Fk:= x1k+x2k+x3k+x4k+x5k+x6k – (y1k+y2k+y3k+y4k+y5k+y6k).  

If Fk = 0 for k = 1,2,3,5,7,9, then (x12+x22+x32+x42+x52+x62)(13F11) = 11F13.


It is readily verified that the first four solns satisfies this.  The pattern is quite obvious and can be extrapolated for systems with powers 11th, 13th, and so on, though a proof is desired that indeed it is generally true.  Second, there may be a partition such that the sum of an equal number of terms on either side is zero, or x1+x2+…+xm = y1+y2+…+ym = 0.  It turns out that for all five solns, it can be done in at least one way.
(Update, 12/21/09):  Wroblewski observed that the conjecture can be generalized as (13F11)(x12+x22+...+x62+y12+y22+...+y62)/2 = 11F13, and similarly for other powers, even if Fk is not zero for k = 2.  Of course, if  it is for k = 2, then it reduces to the form given above.  (End update.)

(Update, 1/13/10):  Wroblewski found the second k = 1,3,5,7,9 using the following special case of a general theorem,


“If [x1, x2, x3, x4, x5]k = [y1, y2, y3, y4, y5]k,   for k = 2,4,6,8,


where x1-x2 = x3-x4 = y1-y2 = y2-y3 = N, then,


[a+x5, -a+y4, -a+y3, a+x1, a+x3, -a+y5]k = [-a+x5, a+y4, a+y1, -a+x2, -a+x4, a+y5]k,  for k = 1,3,5,7,9


where a = N/2.”


Only one known k = 2,4,6,8 has terms with these conditions (see Eighth Powers for details) and yields,


[-767, 399, -299, 995, 1167, -1205] = [-1293, 925, 1279, -57, 115, -679]




If the generic derived 9th degree system is arranged in the above manner, and with terms labeled as {pi, qi}, the form will satisfy the ff five symmetric constraints,


p1+p2 = q1+q2

p1+p6 = q1+q6

p4-p5 = q4-q5

p1+p3+p5 = q1+q3+q5

p2+p4+p6 = q2+q4+q6


Together with k = 3,5,7,9 (since k = 1 is implied), this is a system of 9 eqns in 11 unknowns (since, by scaling, one term can be set equal to unity without loss of generality.)  This can be completely solved as,


(1+a)k + (-1+b)k + (-3+c)k + (2+f)k + (2+e)k + (-1+d)k =

(-1+a)k + (1+b)k + (3+c)k + (-2+f)k + (-2+e)k + (1+d)k


for k = 1,3,5,7,9 and two free variables {u,v} where,


a = 4(u+v),

b = 4(u-v)

2c2 = -3+t+31uv-3w

2d2 = 53+t+71uv-11w

{e2, f2} = {p-q,  p+q}




p = (1/2)(13+t+9uv-5w)

(1/2)q2 = 20+4t+301uv+35u2v2-5(4+11uv)w


where w2 = -15+2t+22uv-135u2v2, and t = 32(u2+v2).


Thus, six variables S = {c,d,e,f,q,w} must be made squares, with {e,f} being two roots of an octic, and ±w is equally valid, though only one sign may yield a rational value.  One non-trivial soln is {u,v} = {46/263,  423/526} which gives the original equation that inspired the form.  Even with six expressions to be made squares, due to the symmetric nature of the system, there is in fact an infinite number of {u,v} such that, with the sign of ±w chosen appropriately, all S are squares, one of which is the family,


v = 4(u-1)/(12u-7)


but unfortunately just yields trivial terms.  (To find two others, one can expand at k = 11.)  However, it may be possible there are other non-trivial solns to this system. (End update.)


(Update, 7/25/09):  Recent work by A. Bremner and J. Delorme ("On Equal Sums of Ninth Powers", Math. of Comp, July 2009) have shown that the multi-grade system,


x1k+x2k+x3k+x4k+x5k+x6k = y1k+y2k+y3k+y4k+y5k+y6k   (eq.1)


for k = 1,2,3,9 has an infinite number of distinct and non-trivial solns.  They solved this by assuming,


{x1, x2, x3, x4, x5, x6} = {u1+w, u2+w, u3+w, v1-w, v2-w, v3-w}

{y1, y2, y3, y4, y5, y6} = {u1-w, u2-w, u3-w, v1+w, v2+w, v3+w}


Note that this satisfies the particular set of constraints,


x1-y1 = x2-y2 = x3-y3 = -(x4-y4) = -(x5-y5) = -(x6-y6) = 2w
and, x1-y4 = -x4+y1;  x2-y5 = -x5+y2;  x3-y6 = -x6+y3.


Eq.1 is identically true for k = 1 while k = 2,3 are also if,


u1n+u2n+u3n = v1n+v2n+v3n,  for n = 1,2   (eq.2)


After finding {ui, vi} such that this is satisfied, one can then find a relationship between them and the free variable w so that it is valid for k = 9 as well.  One can use two approaches:


Method 1.  Bremner-Delorme




{u1, u2, u3} = {a-b+c, c+bt, a+c+at}

{v1, v2, v3} = {a-b+c+at, a+c+bt, c}


These make eq.1 valid for k = 1,2,3.  In the paper, after a lot of clever math, Bremner and Delorme gave the explicit xi, yi.  This author started with those and derived the {ui, vi} given above.  Expanding for k = 9, one gets a quintic (and some trivial linear factors) in {a,b,c}, but a quartic in w.  One can factor this polynomial into a quadratic and cubic using the relation,


w = 2a-b+3c+(a+b)t


The quadratic can be disregarded as it yields only complex solns.  The cubic factor, rather complicated to explicitly write here, is homogeneous in {a,b,c} and can be treated as an elliptic surface such that from an initial soln, one can compute an infinite more.  One set among many given by Bremner and Delorme is,


{a,b,c,t} = {-1, 8 , -5, 3}


and with w = -4, this yields,


[-18, 15, -13, -13, 22, -1] = [-10, 23, -5, -21, 14, -9],  for k = 1,2,3,9.


Note:  For convenience, this author changed the paper's variables from {q1, q2, q4} to {-a, b, c}.  However, values {a,b,c,t} such that w = 0 are trivial.


Method 2.  Piezas


An alternative way is that k = 9 becomes a quartic that factors into two quadratics.  Eq.2 can be completely solved by,


{u1, u2, u3} = {m-pr+q,  -r-mp+q,  pr+q+r}

{v1, v2, v3} = {m+pr+q,  r-mp+q,  -pr+q-r}


Note that the r in the ui are just negated in the vi.  (This complete parametrization was also used by this author for the multi-grade system [k,3,3] for k = 1,2,6.)  Expanding eq.1 for k = 9 results in a quartic in r with only even exponents plus the trivial linear factors,


(1+p)(m+r)(m-r)prw = 0    (eq.3)


To factor the quartic, use the relationship,


m(p-1)-3q+w = 0


Solve for q and substitute when k = 9.  One quadratic factor in r (call this Q1) can be disregarded since, as before, it only yields complex solns.  The other one (call this Q2), for rational solns, its discriminant D which is only a quartic polynomial must be made a square.  For convenience, set w = n/2 and we find this as the rather simple form,


D: = f2m4-2efm3n-3e2m2n2-7fmn3-7en4 = y2


where {e, f} = {(7/2)(p2+p+1),  2(p-1)(p+2)(2p+1)}


Treating D as an elliptic "surface", given an initial soln, one can then find an infinity, as in the previous method.  (In fact, the relationship given above was tediously derived from that method.)  Thus, any rational value of m,n,p such that y is rational but does not involve the trivial factors of eq.3, then non-trivially solves eq.1 for k = 1,2,3,9.  An example is {p,n} = {-1/4, -2} which, after letting {m, r} = {a, b} for aesthetic reasons, gives {xi, yi} as,


xi = {-16+7a+3b,  -16-2a-12b, -16-5a+9b;  8+7a-3b,  8-2a+12b,  8-5a-9b}

yi = {8+7a+3b,  8-2a-12b,  8-5a+9b;  -16+7a-3b,  -16-2a+12b,  -16-5a-9b}


(Note that the b's in the xare just negated in the yi.)  Expanding for k = 9, one must solve the quadratic,


-128-39a2+5a3-9(13+5a)b2 = 0


The discriminant D of this (after removing a square numerical factor) is,


D: = (13+5a)(-128-39a2+5a3) = y2


which, if to be made a square, is an elliptic curve.  Two small solns are a = {-5, -7/2}.  From these initial values, one can then compute more.  These give, up to permutation, the same xi, yi in the first method.  Using the original variables, finding w from n, and deriving r from Q2, we have all the unknowns as {m,p,q,r,w} = {-5, -1/4, 7/4, 4, -1} and get an initial soln to eq.1 as [-13, -18, 15, -13, 22, -1] = [-5, -10, 23, -21, 14, -9],  for k = 1,2,3,9.  Using the other rational points on this elliptic curve will then yield an infinite number of solns to eq.1.


Note 1:  Of course, other choices of {p,n} will give different elliptic curves.

Note 2:  Whether in the first or second method, one can extend the range of exponents up to k = 1,2,3,5,9, but all solns, unfortunately, are now complex.  To cover k = 7 as well did not yield any non-trivial linear relationship.

Note 3:  This author is of the opinion that with a different set of side conditions, it will be possible to find an identity, whether as an elliptic curve or as polynomials, for the multi-grade system k = 1,3,5,7,9, just like its higher counterpart k = 2,4,6,8,10.  (End update.)
9.4  Fourteen Terms
(No identities are yet known.)
9.5  Sixteen Terms
(Update, 2/15/10): Wroblewski, Piezas


A multi-grade [k.4.4] for k = 2,4 can be used to find a nonic [k.8.8] for k = 1,2,3,4,5,9.  Given six variables {p,q,s,t,y,z},


System 1:

[10p+2q-s+t, -10p-2q-s+t, 2p-10q+y-z, -2p+10q+y-z]k =

[10p-2q-s+t, -10p+2q-s+t, 2p+10q+y-z, -2p-10q+y-z]k,  for k = 2,4


System 2:

[5p+q+s, -5p-q+s, 5p+q+t, -5p-q+t,  p-5q+y, -p+5q+y, p-5q+z, -p+5q+z]k =

[5p-q+s, -5p+q+s, 5p-q+t, -5p+q+t , p+5q+y, -p-5q+y, p+5q+z, -p-5q+z]k,  for k = 1,2,3,4,5


is identically true using the substitutions,


{p,q,s,t,y,z} = {u/2, v/2, a+b, a-b, a+c, a-c}

{b,u,c,v} = {eg-2fh, eh+fg, eg+2fh, eh-fg} 


for arbitrary {a,e,f,g,h}.  It can be shown that [2] is derived from [1] using Theorem 5 of the Prouhet-Tarry-Escott problem.


A) One can make the first system valid for k = 6 and the second for k = 6,7 as well if,


(10g2-13h2)e2-(13g2-40h2)f2 = 0


for arbitrary a.  This is reducible to an elliptic curve, with one soln as {e,f,g,h} = {171, 97, 7, 47}, and an infinite more can then be computed. 


B) However, if System 2 is to be true only for k = 1,2,3,4,5,9, then one must solve a quadratic in a with a sextic discriminant that must be made a square.  The only known soln found by Wroblewski is {a,e,f,g,h} = {104, 7, 3/2, -18, 17}, though others may exist.  Whether with the right constraints this can be reduced to a quartic polynomial to be made a square, hence an elliptic curve, is unknown. (End update)

(Update, 2/16/10): As observed by Bremner and Delorme, the smallest [k.6.6] for k = 9, found by Lander et al through computer search back in 1967, is in fact good for k = 1,2,3,9,


[18, 23, 13, -10, -15, -5]k = [21, 22, 9, -13, -14, -1]k


Notice also how appropriate pairs all have the same sum,


18-10 = 23-15 = 13-5 = 21-13 = 22-14 = 9-1 = 8


This “numerical curiosity” has an explanation, which lies in the simple eqn,


142 + 192 + 92 = 172 + 182 + 52


and is just a special instance of Theorem 5 of the Prouhet-Tarry-Escott Problem.


Define, Fn = an+bn+cn+dn-(en+fn+gn+hn) and the [k.8.8],
[x+a, x+b, x+c, x+d, x-a, x-b, x-c, x-d]k = [x+e, x+f, x+g, x+h, x-e, x-f, x-g, x-h]k


1) If F2 = 0, and a variable x such that 42F4x4+28F6x2+3F8 = 0, then the system is for k = 1,2,3,9. 


2) If F2 = F4 = 0 and 28F6x2+3F8 = 0, then it is for k = 1,2,3,4,5,9.


Notice how appropriate pairs of terms all have the same sum 2x.  Also, it reduces to a [k.6.6] if a pair from opposite sides, say d = h = 0.  Lander’s soln was simply the case,


{a,b,c,d} = {14, 19, 9, 0}

{e,f,g,h} = {17, 18, 5, 0}


and x = 4.  Bremner and Delorme found a relationship between the terms such that the quartic in x factored into two quadratics.  Making its discriminant a square involved an elliptic curve, so there are an infinite number of solns to [1].  This is discussed more in Section 9.3 above.


For non-zero {d,h}, no example is yet known for [1], but for [2], Wroblewski directly found a nonic [k.8.8] for k = 1,2,3,4,5,9, which yields,


{a,b,c,d} = {240, 63, 197, 122}

{e,f,g,h} = {167, 10, 168, 243}

and x  = 52.  Since, these variables can be reduced in size with appropriate transformations given in the previous update, it is possible there are others. (End update.)

(Update, 11/4/15):  Piezas


Using a variant of Theorem 5 discussed in Equal Sums of Like Powers, we can use an 8th power [k,5,5] identity to get a 9th power [k,8,8] 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+2c, 3b+c+d, 3b+c-d, 4a+c, 3a-2c,  b-c+d, b-c-d,  4b-c]k =

[a-2c,  3b-c+d,  3b-c-d,  4a-c, 3a+2c, b+c+d, b+c-d, 4b+c]k


for k = 1,2,3,5,7,9, and same conditions a2+12b2 = c2, 12ba2+b2 = d2.  Notice that c has just been negated in the RHS.  (End update.) 



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