030: Tenth Powers

(8b)^k + (5a-4b)^k + (-a-2d)^k + (a-2d)^k + (-5a-4b)^k + (a-4b)^k + (-12b+4c)^k + (12b+4c)^k =

(4a+8b)^k + (a-4b)^k + (3a-2d)^k + (-3a-2d)^k + (-4a+8b)^k + (-16b)^k + (a+4c)^k + (-a+4c)^k

With terms as {x_i, y_i}, the identity obeys the constraints,

x₁-y₁ = -(x₂-y₂) = x₃-y₃ = -(x₄-y₄) = -4a

x₅-y₅ = -(x₆-y₆) = x₇-y₇ = -(x₈-y₈) = -(a+12b)

(Update, 1/11/10)

10.3 Fourteen Terms

The smallest soln to,

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k+x₆^k+x₇^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k+y₆^k+y₇^k (eq.1)

for k = 10 is,

[1, 8, 15, 26, 26, 33, 38]¹⁰ = [22, 23, 24, 29, 32, 35, 36]¹⁰

given by Ekl (“New results in equal sums of like powers”, 1998). However, as far back in 1939, Moessner already managed to solve eq.1 (pre-computer searching!). The second(?) smallest soln, and the smallest in distinct integers, turns out to be multigrade,

[32, -1, -61, -55, -31, -13, -28, 68]^k =

[44, -13, -49, -67, 20, -64, 23, 17]^k

for k = 1,2,4,6,8,10, and belongs to an infinite family. Note that its terms obey the relations,

-32+44 = -1+13 = 61-49 = -55+67 = 12

31+20 = -13+64 = 28+23 = 68-17 = 51

Wroblewski, Piezas

Let a²+b² = c² and a²+52b² = d², a pair of conditions which is a concordant form. Then a (k,7,7) for k = 1,2,4,6,8,10 is,

(Update, 1/12/10): The identity has the form,

[pa+qb+c, pa+qb-c, qa-pb+d, qa-pb-d, ra+sb, sa-rb]^k =

[pa-qb+c, pa-qb-c, qa+pb+d, qa+pb-d, ra-sb, sa+rb]^k

where ma²+nb² = c² and na²+mb² = d². Notice how the variable b is just negated in the RHS. Wroblewski did a search and found two new solns, though only for k = 2,4,6,8:

{p,q,r,s; m,n} = {1, 3, 2, 8; 45, -11}; for k = 2,4,6,8,10

{p,q,r,s; m,n} = {2, 5, 4, 6; 64/5, -11/5}

{p,q,r,s; m,n} = {1, 10, 1, 11; 248/5, -27/5}

See also Eighth Powers. (End update.)

for k = 10 has relatively small solns. The first one was found by Randy Ekl in 1997,

[5, 23, 34, 34, 85, 92] = [16, 25, 28, 32, 71, 95]

There are many more and whether these belong to a parametric family is unknown. However, a family has been found for the even multi-grade system k = 2,4,6,8,10 (pls see update below). A numerical example (not belonging to the known family) was first found in 1999,

{22, 61, 86, 127, 140, 151} = {35, 47, 94, 121, 148, 146}

by Kuosa and Meyrignac as k = 10, and Shuwen (who noticed it was not just for k = 10 but was multi-grade) which automatically leads to an ideal soln of deg 11. To see if this is also for k = 1 if the terms are signed, there are now 64 possible sums of ± a ± b ± c ± d ± e ± f (and an equal number for the terms on the other side), and Shuwen found that sums of one side are equal to the other in seven ways, three of which are symmetric,

a-b-c+d-e+f = g-h-i+j-m+n = 13

a+b+c-d-e+f = g+h+i-j-m+n = 53

a+b-c-d+e+f = g+h-i-j+m+n = 161

a+b+c-d+e+f = -g+h-i+j+m+n = 333

a+b-c+d-e+f = -g-h+i+j-m+n = 135

a-b+c+d-e+f = -g-h+i-j+m+n = 185

a-b+c-d+e-f = g-h+i+j-m-n = -91

not counting seven other combinations which simply negate the terms. Also, similar to the known family, it is also the case that,

x₃²+x₄² = y₂²+y₆²

Whether this is: (1) a fluke, or (2) a property of the possible family which this example belongs to, or (3) solns to even systems with signed terms tend to be valid for k = 1 the more terms there are, is not known. While there are parametric (k.6.6) with all three side-conditions,

x₁ ± x₂ = y₁ ± y₂

x₃ ± x₄ = y₃ ± y₄

x₅ ± x₆ = y₅ ± y₆

for sextic and octic systems, it is not known if it can be for the decic. (The Choudhry-Wroblewski decic family uses only two.)

Note: Kuosa and Meyrignac were looking at 10.6.6 and among the 14 solns they found, one turned out to be for k = 2,4,6,8,10. If this was just a statistical fluke, one might expect some of the others to be also valid for, say, k = 2 or 4. This author checked and none is good for any other even exponent. So there is the situation of a small soln that is suddenly multi-grade for six exponents, counting k = 1. Together with its other properties, it seems there might be more to it than meets the eye.

Update (June 6, 2009): Choudhry, Wroblewski

It has been brought to my attention (by J. Wroblewski) that the case of an equal sum of six tenth powers,

a₁^k+a₂^k+a₃^k+a₄^k+a₅^k+a₆^k = b₁^k+b₂^k+b₃^k+b₄^k+b₅^k+b₆^k, for k = 2,4,6,8,10

has already been cracked (in 2008) and shown to have an infinity of solns by reducing it, like the other systems, to a particular elliptic curve, though the Kuusa-Meyrignac-Shuwen soln does not belong to this family. (See their paper "Ideal Solutions of the Tarry-Escott Problem of Degree Eleven with Applications to Sums of Thirteenth Powers", Hardy-Ramanujan Journal, Vol. 31, 2008.) Choudhry and Wroblewski defined a_i and b_i as,

a₁ = 2xy+x+2y-7; b₁ = 2xy+2x+y-7

a₂ = 2xy-x-2y-7; b₂ = 2xy-2x-y-7

a₃ = 2xy-2x+y+7; b₃ = 2xy-x+2y+7

a₄ = 2xy+2x-y+7; b₄ = 2xy+x-2y+7

a₅ = 3x+5y; b₅ = 5x+3y

a₆ = 5x-3y; b₆ = 3x-5y

where a variable z has been set z = 1 without loss of generality. Substituting these values into the system, the higher eqns hold if the ff condition is satisfied,

8x²y²-17x²-17y²+98 = 0

Do the simple change of variables {x, y} = {u, v/(8u²-17)} and the condition reduces to,

(-17+8u²)(-98+17u²) = v²

or a quartic polynomial in u that is to be made a square. As Choudhry and Wroblewski proved, this is an elliptic curve. One trivial soln is u = 1 but from this initial point, we can compute an infinite number of non-trivial ones such as u = -457/353, etc. However, by looking at the solns a_i and b_i, one can see certain relationships and symmetries between them. Using those relationships, this author will show that it can be simplified to a form analogous to the Letac-Sinha identity for k = 2,4,6,8.

(Update, 2/1/10): Choudhry, Wroblewski

The 6th power (k.4.4) identity can be generalized to the form (k.6.6) by adding the italized pair of terms,

[-(axy+bx+cy-d), axy-bx-cy-d, axy-cx+by+d, -(axy+cx-by+d), ex+fy, fx-ey]^k =

[-(axy+by+cx-d), axy-by-cx-d, axy-cy+bx+d, -(axy+cy-bx+d), ey+fx, ex-fy]^k

where again, {x,y} merely swap places. (Since the eqn is homogeneous, one may assume the leading coefficient of the terms as a = 1 without loss of generality.) This obeys the same basic constraints (up to sign changes),

p₁-p₂ = q₁-q₂,

p₃-p₄ = q₃-q₄,

(-p₁)^k+p₂^k+p₃^k+(-p₄)^k = (-q₁)^k +q₂^k+q₃^k+(-q₄)^k, k = 1,2

p₅²+p₆² = q₅²+q₆²

If we assume,

p₁+p₂+p₅ = q₁+q₂+q₅,

p₃+p₄+p₆ = q₃+q₄+q₆,

then it must be the case that 2(b-c) = e-f (eq.1). Choudhry and Wroblewski found for k = 2,4,6,8,10, the equivalent non-trivial solns,

{a,b,c,d,e,f} = {1, 1, 2, 14, 3, 5}, where 2x²y²-17(x²+y²)+392 = 0

{a,b,c,d,e,f} = {2, 1, 2, 7, 3, 5}, where 8x²y²-17(x²+y²)+98 = 0

Disregarding eq.1, two other distinct non-trivial solns found (by Piezas), though only up to k = 2,4,6,8, are,

{a,b,c,d,e,f} = {5, 3, 7, 3, 2, 10}, where 125x²y²-53(x²+y²)+45 = 0

{a,b,c,d,e,f} = {5, 9, 11, 11, 10, 12}, where 125x²y²-221(x²+y²)+605 = 0

though this author is not certain if these are the same 8th power multi-grades found by Wroblewski. See also Sixth Powers for the context of this identity. (End update.)

Piezas

(a+3b+c)^k+(-a-3b+c)^k+(3a-b+d)^k+(-3a+b+d)^k+(2a+8b)^k+(-8a+2b)^k =

(-a+3b+c)^k+(a-3b+c)^k+(3a+b+d)^k+(-3a-b+d)^k+(-2a+8b)^k+(-8a-2b)^k

for k = 2,4,6,8,10, where 45a²-11b² = c² and -11a²+45b² = d².

where the ratio a/b = {2, 1/2, 3/2, 2/3} must be avoided as it yields trivial solns. Labelling terms as a_i and b_i, then a small tweak in the sign of {a₁, a₄, b₁, b₄} can make it valid for k = 1 as well. Note that, just like in the Letac-Sinha identity, the two quadratic polynomials to be made squares have palindromic or reversible coefficients. One can easily verify this soln by solving {c,d} in radicals for arbitrary {a,b} and substituting them into the system. Of course, if they are to be rational, then appropriate {a,b} must be chosen, with the smallest non-trivial one being {a,b} = {186, 331} giving,

[886, -293, 1180, 953, 1510, -413] = [700, -107, 1511, 622, 1138, -1075]

after removing a small common factor. By solving the first condition as a quadratic form {a, b} = {u²+11v², 2u²+2uv-22v²} using this on the other will result in a quartic polynomial that is to be made a square, with some {u,v} as trivial, but a non-trivial is {u,v} = {12, -5}. Treating this as an elliptic curve, from this initial point, an infinite number of rational solns can then be computed. This identity was found using the relationships between the a_i and b_i and, whether in the Choudhry-Wroblewski version or by this author, they satisfy seven side conditions:

a₁+a₂ = b₁+b₂

a₃+a₄ = b₃+b₄

a₁-a₂-a₅ = b₁-b₂-b₅

a₃-a₄+a₆ = b₃-b₄+b₆

a₁+a₂+a₃-a₄+a₅+a₆ = b₁+b₂-b₃+b₄+b₅-b₆

a₁^k+a₂^k+a₃^k+a₄^k = b₁^k+b₂^k+b₃^k+b₄^k, for k = 1,2

and together with the five eqns,

a₁^k+a₂^k+a₃^k+a₄^k+a₅^k+a₆^k = b₁^k+b₂^k+b₃^k+b₄^k+b₅^k+b₆^k, for k = 2,4,6,8,10

we have 12 (or 11) eqns in 12 unknowns, call this augmented system as S₁₀₊. By solving this system, it can be shown that the identity given by this author is the only non-trivial soln to S₁₀₊. (There may be just 11 eqns as one may be a consequence of the others.)

Note 1: By negating {a₁, a₄, b₁, b₄}, the identity in fact is also valid for k = 1. Also, by eliminating or changing one of the side conditions, it may be possible to come up with a different identity from this one. What would be an appropriate change, I do not yet know.

Note 2: I guess my postscript at the final section need to be modified. (However, the case k = 1,3,5,7,9 still needs to be cracked.)

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k+x₆^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k+x₆^k (eq.1)

10.2 Twelve Terms

The eqn,

x₁¹⁰+x₂¹⁰+x₃¹⁰+x₄¹⁰+x₅¹⁰ = y₁¹⁰+y₂¹⁰+y₃¹⁰+y₄¹⁰+y₅¹⁰

10.1 Ten terms

10.2 Twelve Terms

10.3 Fourteen Terms

10.4 Sixteen Terms

10.5 Twenty Terms

10.1 Ten Terms

No non-trivial solution in the rationals is yet known to equal sums of five 10th powers,