Updates Page 01: Jan
(Update, 1/11/10)
10.3 Fourteen Terms
Wroblewski, Piezas
Let 4a2+13b2 = c2 and 16a2+b2 = d2. Then for k = 1,2,4,6,8,10,
(2b)k + (-5a-b)k + (5a-b)k + (-a-c)k + (a-c)k + (-3b+d)k + (3b+d)k =
(-4b)k + (-4a+2b)k + (4a+2b)k + (-a+d)k + (a+d)k + (-3a-c)k + (3a-c)k
though trivial ratios such as a/b = {1, 3, 3/5} should be avoided as the system becomes trivial. A non-trivial example is {a,b} = {3, 16}. As the two quadratic conditions define an elliptic curve, then there is an infinite number of solns. This is a special case of a more general identity involving sixteen terms.
10.4 Sixteen Terms
Piezas, Wroblewski
The latter author suggested the system,
[-a-c+x, a+c+x, -b+c+z, b-c+z, a-b-c-y, -a+b+c-y, -a-b-c-t, a+b+c-t]k =
[-a+c+x, a-c+x, b+c+z, -b-c+z, -a-b-c-y, a+b+c-y, -a+b-c-t, a-b+c-t]k
for k = 1,2,4,6,8,10, and the former found the complete soln using Mathematica. This entails making four quartic polynomials as squares. Set a = 1 without loss of generality and define {x,y,z,t} = {√p, √q, √r, √s}. Then {p,q,r,s} can be rationally expressed in terms of the free variables {b,c}. As the actual expressions are unwieldy, first define,
p-s = (-2b+9b2+8c+2bc-8c2)/3
q-s = 8(1-b)(1-c)(-b+c)c/(3b+3c+3bc)
q-r = (-9+2b-2c-8bc+8c2)/3
and it is easy to solve for {p,q,r}. Substituting these into the system and factoring at k = 6,8,10, one will find a common linear factor for the last unknown s. Expanding for k = 12 will yield the trivial forms of {b,c}. A non-trivial example which makes {p,q,r,s} squares is {b,c} = {17/12, 5/3}.
However, Wroblewski found that if b+c = 1/2, then {p,s} are squares, {q,r} become just quadratic polynomials, and two terms {x8, y1} are identical, so it reduces to the simpler one in section 10.3. (End update)
(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. 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 = 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. If indeed first solns are relatively small, it may be feasible to find the one for k = 13.
II. Twelfth Powers
12.1 Twelve Terms
(No solution is yet known.)
12.2 Fourteen Terms
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 a single numerical example for k = 12. "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 trend continues, the first soln might have terms just in the lower end of 3 digits. (End update.)
[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)
[57, 679, 1293, -115, -925, -1279] = [299, 767, 1205, -399, -995, -1167]
[407, 163, 341, -37, -371, -119] = [221, 311, 403, -23, -181, -347]
(Update, 1/11/10): For twelve terms, or an equal sum of sixth ninth 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]
(Update, 1/17/10): Wroblewski
A special case of the Birck-Sinha Theorem (with some terms changed in sign) is,
[a+p, a-p, d+p, d-p, b+c, b-c, 2q]k = [b+q, b-q, c+q, c-q, a+d, a-d, 2p]k
for k = 2,4,6,8 where {c,d} are,
a2+3p2+q2 = c2
b2+p2+3q2 = d2
and aq = bp. For the special case when we equate two terms, one from each side, as 2p = a-p, four terms cancel out hence this reduces from a (k.7.7) to the (k.5.5) Birck-Sinha Identity discussed in a previous section. (End update.)
(Update, 1/17/10) Wroblewski added four more terms to each side of the 12-term form to get,
[pa+qb+c, pa+qb-c, qa-pb+d, qa-pb-d, ra+sb, sa-rb, ta+ub, ua-tb, va+wb, wa-vb]k =
[pa-qb+c, pa-qb-c, qa+pb+d, qa+pb-d, ra-sb, sa+rb, ta-ub, ua+tb, va-wb, wa+vb]k
for k = 2,4,6,8,10, where ma2+nb2 = c2 and na2+mb2 = d2. Again, the variable b is just negated in the RHS. Wroblewski found three non-trivial solns,
{p,q,r,s,t,u,v,w; m,n} = {4, 4, 3, 11, 5, 7, 3, 7; 45, -11}
{p,q,r,s,t,u,v,w; m,n} = {4, 4, 9, 17, 15, 13, 13, 7; 189, 85}
{p,q,r,s,t,u,v,w; m,n} = {3, 3, 11, 17, 16, 14, 14, 10; 220, 136}
though certain ratios a/b must be avoided and which can determined by expanding for k = 12. Interestingly, the first soln uses the same {m,n} as the one for 12 terms! As these two quadratic conditions define an elliptic curve, it is unknown why certain curves are "favored", appearing again and again (it also appears in a (k.4.4) for k = 1,2,4,6). It is also remains to be known whether: a) there are others, b) if there is w = v = 0 so reduces to 16 terms, or c) if it can be non-trivially valid for k = 12. (End update)
(Update, 1/13/10): Wroblewski
Using a special case of a general theorem, Wroblewski used a k = 2,4,6,8 to derive a k = 1,3,5,7,9!
“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.”
It can be seen that the second eqn (or eq. 2) found by Borwein, Lisonek, Percival satisfies this and thus yields,
[-767, 399, -299, 995, 1167, -1205] = [-1293, 925, 1279, -57, 115, -679]
which is the second k = 1,3,5,7,9 since the first one was found in 2000 by Shuwen via computer search. Since then, Wroblewski has found a few others also by computer search. See Ninth Powers for more details. (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]
Piezas
By arranging the generic derived 9th degree system 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+d)k + (2+e)k + (-1+f)k =
(-1+a)k + (1+b)k + (3+c)k + (-2+d)k + (-2+e)k + (1+f)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
{d2, e2} = {p+q, p-q}
2f2 = 53+t+71uv-11w
and
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 squares, 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, 1/12/10): Piezas, Wroblewski
Starting with the known soln for k = 2,4,6,8,
[-508, -245, 331, 18, 471] = [-366, 189, -103, 452, 515]
Wroblewski suggested the general form,
[x, x+a+b, x+b+d, x+2a+2b, p]k = [x+b, x+d, x+a+2b, x+a+b+d, q]k
where, for this case, {a,b,d,p,q,x} = {121, 142, 697, 471, 515, -508}. This system can be resolved into a final eqn that is only a quartic, though the expressions are messy. The first author translated the general form into a more symmetric version and found the complete soln in terms of four quadratics to be made squares. Let,
[a+b, -a+b, a+b+2c, a+b-2c, e]k = [b-c-d, b+c-d, b-c+d, b+c+d, f]k (eq.1)
for k = 2,4,6,8, then with {a,b}as free variables, {c,d,e,f} are given as,
32c2 = 5a2-10ab+8b2
32d2 = 29a2+6ab+72b2
4e2 = 9a2-6ab+36b2
4f2 = 13a2+2ab+4b2
with some {a,b} as trivial like the ratio a/b = {2, 6, etc). A non-trivial soln is {a,b} = {-288, 43} which makes all {c,d,e,f} as squares and yields the multi-grade which inspired this form. It is easy to make {c,d} as squares using an elliptic curve, though apparently it is only by chance that {e,f} are squares as well. But this can provide a rational family to the system,
x1k + x2k + x3k + x4k + x5k/2 = y1k + y2k + y3k + y4k + y5k/2, for k = 2,4,6,8 (End update.)
(Update, 1/12/10): II. x1k+x2k + … + x6k = x1k+x2k + … + x6k , for k = 2,4,6,8
Wroblewski gave five new beautiful identities:
1. If 64a2-11b2 = 5c2 and -11a2+64b2 = 5d2, then
(2a+5b+c)k + (2a+5b-c)k + (5a-2b+d)k + (5a-2b-d)k + (4a+6b)k + (6a-4b)k =
(2a-5b+c)k + (2a-5b-c)k + (5a+2b+d)k + (5a+2b-d)k + (4a-6b)k + (6a+4b)k
for k = 2,4,6,8. Trivial ratios are a/b = {1, 2}. These two quadratic conditions define an elliptic curve, hence the system has an infinite number of rational solns.
2. If 248a2-27b2 = 5c2 and -27a2+248b2 = 5d2, then
(a+10b+c)k + (a+10b-c)k + (10a-b+d)k + (10a-b-d)k + (a+11b)k + (11a-b)k =
(a-10b+c)k + (a-10b-c)k + (10a+b+d)k + (10a+b-d)k + (a-11b)k + (11a+b)k
for k = 2,4,6,8. Trivial ratios are a/b = {1, 3}, while a non-trivial soln is {a,b} = {9533, 3439} from which an infinite more can be computed. These belong to the general 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 which first appeared in Tenth Powers by this author. Notice how the variable b is just negated in the RHS. Thus, there are now three known non-trivial solns,
{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}
and whether there are more is unknown, though Wroblewski has done a numerical search for {p,q,r,s} within a bound. (By expanding the system, one can linearly express {m,n} in terms of {p,q,r,s}, so the latter are the true unknowns.)
3. If -15a2+96b2 = 9c2 and -96a2+825b2 = 9d2, then
(a+8b+c)k + (a+8b-c)k + (2a-b+d)k + (2a-b-d)k + (3a-2b+c)k + (3a-2b-c)k =
(a-8b+c)k + (a-8b-c)k + (2a+b+d)k + (2a+b-d)k + (3a+2b+c)k + (3a+2b-c)k
for k = 2,4,6,8. Trivial ratios are a/b = {1, 2, 3, 5/2}.
This generalizes the form above into,
[pa+qb+c, pa+qb-c, ra+sb+d, ra+sb-d, ta+ub+c, ta+ub-c]k =
[pa-qb+c, pa-qb-c, ra-sb+d, ra-sb-d, ta-ub+c, ta-ub-c]k
where p1a2+p2b2 = c2, and p3a2+p4b2 = d2. Again, the variable b has simply been negated in the RHS. Whether there are other non-trivial solns of this form, or if it can be extended up to k = 10, is unknown.
4. If 17a2-33b2 = 8c2 and 3a2-3b2 = 8d2, then,
(a+b)k + (a-b)k + (a+c)k + (a-c)k + (b+3d)k + (b-3d)k =
(b+c)k + (b-c)k + (b+d)k + (b-d)k + (2a)k + (4d)k
for k = 2,4,6,8. Trivial ratios are a/b = {1, 5, 7/5}. This is a different form which implies there may be many others just waiting to be discovered.
5. Given,
2(a2+b2+c2) = m2+11n2 (eq.1)
8(a4+b4+c4) = m4+54m2n2+89n4 (eq.2)
then, (2a)k + (2b)k + (2c)k + (m+n)k + (m-n)k + (4n)k = (a+b+c)k + (a-b+c)k + (a+b-c)k + (a-b-c)k + (m+3n)k + (m-3n)k, for k = 2,4,6,8.
Note that the eqn is already identically true for k = 2 since,
(2a)2 + (2b)2 + (2c)2 = (a+b+c)2 + (a-b+c)2 + (a+b-c)2 + (a-b-c)2
(m+n)2 + (m-n)2 + (4n)2 = (m+3n)2 + (m-3n)2
(See the Birck-Sinha theorem below.) However, the relation 2a±m±3n = 0 must be avoided as it is trivial. Wroblewski found several non-trivial linear relations between the {a,b,m,n},
m = a+b
m = 2(a+b)/3
m = 3a+2b
2a+2b = m+n; {a,b,m,n} = {d+u+v, -d+u+v, 4u, 4v}
2a+2b = m+5n; {a,b,m,n} = {d+u+5v, -d+u+5v, 4u, 4v}
such that for [1], [2], [3], {c2, n2} are quadratics in {a,b}; while for [4], [5], {c2, d2} are quadratics in {u,v}, hence define an elliptic curve. For example, given [4], then eq.1 and eq.2 are true if,
76u2+88uv+12v2 = 3c2
91u2-50uv+3v2 = 3d2
Piezas found another linear relation,
6. n = (a+b)/3
such that eq.1 and eq.2 are true if,
11a2+10ab+11b2 = 3c2
73a2+38ab+73b2 = 9m2
Q. Any others? For the latest data on higher powers with a restricted number of terms, see Wroblewski's "A Collection of Numerical Solutions of Multi-grade Equations". (End update.)
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(Update, 1/30/10): Choudhry
Form: x18+x28+...+x68 – (y18+y28+...+y68) = N
Choudhry's Eighth Powers Theorem: “Any non-zero rational number N is the sum and difference of twelve rational 8th powers in an infinite number of non-trivial ways."
Proof: Choudhry’s method is equivalent to solving the equation,
U1 – U2 = M
U1:= (ax+y)8 + (p18+p38)(x+by)8 + (p28+p48)(cx-y)8 + (x-dy)8
U2:= (ax-y)8 + (p18+p38)(x-by)8 + (p28+p48)(cx+y)8 + (x+dy)8
for variables {a,b,c,d} and {p1, p2, p3, p4} such that, when expanded and collecting {x,y}, all terms vanish except M which is only linear in x. He found a wonderfully simple soln given by,
{p1, p2, p3, p4} = {t, t3, t5, t7}
{a,b,c,d} = {t12, t12, t4, t20}
such that,
M = 24 t8(t4-t20-t52+t84+t116-t132) xy7
for arbitrary {t,x,y}. Note how the exponents match up as 1+7 = 3+5 = 8, 12+12 = 4+20 = 24, as well as 4+132 = 20+116 = 52+84 = 136. (Q. What is reason for the simplicity of the soln, and can this be generalized?) Doing the substitution {x,y} = {24N, t4-t20-t52+t84+t116-t132}, we then get the form,
M = 28Q8 N
Dividing by the factor, then any non-zero N is the sum and difference of twelve rational 8th powers in an infinite number of ways as claimed. (Using another approach, it may be possible to reduce this to ten or even eight rational 8th powers.) Of course, if N is itself an 8th power, then this provides a polynomial identity for the 13-term,
x18+ x28+… x68 = y18+ y28+… y78
Note: My thanks to Choudhy for providing a copy of his paper "On Sums of Eighth Powers", Journal of Number Theory, Vol 39, Sept 1991. (End update.)
(Update, 1/27/10): Piezas
It can be shown that the system,
[x1, x2, x3, x4, x5]k = [y1, y2, y3, y4, y5]k, for k = 2,4,6,8,
with the constraint x1-x2 = x3-x4 = y1-y2 = y2-y3 can be completely solved in terms of nested square roots. The soln is,
(1-e)k + (-1-e)k + (1-d)k + (-1-d)k + ak = (2-c)k + (-c)k + (-2-c)k + bk + fk
a = 4(u+v),
b = 4(u-v)
2c2 = -3+t+31uv-3w
{d2, e2} = {p+q, p-q}
2f2 = 53+t+71uv-11w
and,
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), for two free variables {u,v}. This then leads to the 9th deg multigrade,
(1+a)k + (-1+b)k + (-3+c)k + (2+d)k + (2+e)k + (-1+f)k =
(-1+a)k + (1+b)k + (3+c)k + (-2+d)k + (-2+e)k + (1+f)k
for k = 1,3,5,7,9 .
Thus, six variables S = {c,d,e,f,q,w} must be squares, 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. If we add one more constraint to the 8th deg system, namely x2-x4 = y2-y4, we lose one free variable but the problem simplifies to four quadratics to be made squares. This is given by,
[b-c-d, b+c-d, -b-c-d, -b+c-d, f]k = [a+b-2c, a+b, a+b+2c, a-b, e]k, for k = 2,4,6,8
where {c,d,e,f} are,
32c2 = 5a2-10ab+8b2
32d2 = 29a2+6ab+72b2
4e2 = 9a2-6ab+36b2
4f2 = 13a2+2ab+4b2
Since the eqns are homogeneous, we can set b = 1 without loss of generality and "a" in fact is just the only free variable. This system has also been investigated by Wroblewski, see below. There are some trivial {a,b} like the ratio a/b = {2, 6, etc). A non-trivial soln is {a,b} = {-288, 43} which makes all {c,d,e,f} as squares and yields the original multi-grade. It is easy to make {c,d} as squares using an elliptic curve, though apparently it is only by chance that {e,f} are squares as well. But this can provide a rational family to,
x1k + x2k + x3k + x4k + x5k/2 = y1k + y2k + y3k + y4k + y5k/2, for k = 2,4,6,8
Considering how this result has a much simpler form, it may be possible that, with the right constraint, the more general case can be simplified still. If we add yet one more condition and require that k = 1 hold as well (for signed terms), then it can be shown that the original eqn is the only soln to this system of 9 eqns (4+5) in 9 unknowns (since one variable can be set equal to 1 without loss of generality). (End update.)
(Update, 1/25/10): Piezas
Given a (k.4.4) for k = 2,4,6, if its terms {xi , yi} are such that x1-x2 = y1-y2 = y2-y3 , then it can lead to a (k.5.5) for k = 1,3,5,7. This has an infinite family,
(3a+2b+c)k + (3a+2b-c)k + (-2a-b+d)k + (-2a-b-d)k = (3a+b+2c)k + (3a+b)k + (3a+b-2c)k + (-7a-b)k, for k = 1,2,4,6
which leads to the higher system,
[3a+b+3c, 3a+2b-2c, -7a-b+c, -2a-b-c-d, -2a-b-c+d]k =
[3a+b-3c, 3a+2b+2c, -7a-b-c, -2a-b+c-d, -2a-b+c+d]k, for k = 1,2,3,5,7
where, for both, 10a2-b2 = c2, and 55a2-6b2 = d2, with trivial ratios a/b = {1, 1/3, 5/13}. It is tempting to speculate that there may be a higher-order version of this for (k.5.5), k = 2,4,6,8, with the form,
[pa+qb+c, pa+qb-c, ra+sb+c, ra+sb-c, ta+wb]k =
[ua+vb+2c, ua+vb, ua+vb-2c, xa+yb+d, xa+yb-d]k (System 3)
where m1a2+m2b2 = c2, and m3a2+m4b2 = d2, for constants {p,q,r,s,t,u,v,w,x,y} and {m1, m2, m3, m4}. System 3 has x1-x2 = x3-x4 = y1-y2 = y2-y3 and would automatically lead to a (k.6.6) for k = 1,3,5,7,9. While there is numerical example with the right constraints, it is unknown if it belongs to a family with an infinite number of rational solns. (End update.)
(Update, 1/25/10): Wroblewski
Given,
(x2)k + (y2)k + (x2+y2)k = (z2)k + (t2)k + (z2+t2)k, for k = 2,4 (eq.1)
or equivalently,
x4+x2y2+y4 = z4+z2t2+t4
then,
[ax+bz, -ax+bz, ay+bt, -ay+bt, at+bx, at-bx, az+by, az-by]k =
[bx+az, -bx+az, by+at, -by+at, bt+ax, bt-ax, bz+ay, bz-ay]k, for k = 1,2,4,8
for arbitrary {a,b}, hence giving an identity in terms of linear forms. An example is {x,y,z,t} = {1, 26, 17, 22}. To find a [8.7.7], since {a,b} are arbitrary, one can equate appropriate xp = yq so a pair of terms cancel out. Note: Eq. 1, or more generally its unsquared version, is at the core of the Ramanujan 6-10-8 Identity. (End update.)
(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.)
(Update, 1/18/10): Wroblewski suggested the family which starts with Pythagorean triples,
(2x)2 + (x2-1)2 = (x2+1)2
2(4x3)4 + 2(2x)4 + (4x4-1)4 = (4x4+1)4
(24x7)8 + 7(23x5)8 + 7(22x3)8 + (2x)8 + (16x8-1)8 = (16x8+1)8
(28x15)16 + 35(27x13)16 + 273(26x11)16 + 715(25x9)16 + 715(24x7)16 + 273(23x5)16 + 35(22x3)16 + (2x)16 + 8(256x16-1)16 = 8(256x16+1)16
However, neat and small decompositions stop at n = 8. (End update)
(Update, 1/18/10): Piezas
Crussol’s soln can be given in a more aesthetic manner as two quadratics to be made squares. Let,
[c+be+af, c-be-af, d+ae-bf, d-ae+bf]k =
[c+be-af, c-be+af, d+ae+bf, d-ae-bf]k
for k = 1,2,4,6, where,
ma2+nb2 = c2 (eq.1)
na2+mb2 = d2 (eq.2)
with {m,n} = {(4e2-f2)/15, (4f2-e2)/15} for some constants {e,f}. These simultaneous equations, eq.1 and eq.2, define an elliptic curve. For ex, let {e,f} = {13, 1} and we get,
(a+13b+c)k + (-a-13b+c)k + (13a-b+d)k + (-13a+b+d)k =
(-a+13b+c)k + (a-13b+c)k + (13a+b+d)k + (-13a-b+d)k
where 45a2-11b2 = c2, and -11a2+45b2 = d2. These conditions are the same as the ones for the only known [k.6.6] identity for k = 1,2,4,6,8,10 discussed in Tenth Powers. More generally, given eq.1 and eq.2, which lead to an elliptic curve E(m,n), it is quite easy to find a k = 1,2,4,6 which uses this curve since the constants {e,f} can be derived from {m,n} as,
{4m+n, m+4n} = {e2, f2}
hence, if the two expressions themselves are squares, then it can be used for a k = 1,2,4,6. (End update.)
(Update, 1/17/10):
8.5 Eighteen terms
Lander
(28k+4 + 1)8 = (28k+4-1)8 + (27k+4)8 + (2k+1)8 + 7((25k+3)8 + (23k+2)8)
Note: This has a counterpart for 4th powers. I misplaced my notes but, if I remember correctly, 7 is replaced by 3(?), and the powers of 2 are changed appropriately. If someone can come up with something, pls submit it.
(Update, 1/17/10): I. x1k+x2k + … + x6k = x1k+x2k + … + x6k , for k = 2,4,8
Wroblewski completely solved the system,
[ap+d, ap-d, bq+c, bq-c, ar, bs]k = [bp+c, bp-c, aq+d, aq-d, as, br]k, for k = 2,4,8
with the palindromic conditions ma2+nb2 = c2 and na2+mb2 = d2 for some constants {p,q,r,s; m,n} as an elliptic curve imbedded in another elliptic curve. The soln is,
{p,q,r,s; m,n} = {u+v, u-v, u-2v, u+2v; w, v2}
where {u,v,w} satisfy the elliptic curve,
3u4+20u2v2+16v4 = w2 (eq.1)
The first point is {u,v,w} = {2,1,12} which yields {p,q,r,s; m,n} = {3,1,0,4 ; 12, 1} and solving the simultaneous quadratics 12a2+b2 = c2 and a2+12b2 = d2 requires another elliptic curve. Since r = 0, this in fact is the known (8.5.5) identity discussed in Ten Terms, hence is valid for k = 2,4,6,8. The second point is {u,v,w} = {6,1,68} which gives {7,5,4,8; 68, 1} so,
[7a+d, 7a-d, 5b+c, 5b-c, 4a, 8b]k = [7b+c, 7b-c, 5a+d, 5a-d, 8a, 4b]k
for k = 2,4,8 where 68a2+b2 = c2 and a2+68b2 = d2. And so on for all rational points of eq.1. (End update.)