PART 11. Sum / Sums of Eighth Powers
8.0 Six terms 8.1 Eight terms 8.2 Nine terms 8.3 Ten terms 8.4 Twelve terms 8.5 Fourteen terms 8.6 Sixteen terms 8.7 Eighteen terms
8.0 Six terms
It is conjectured that,
x_{1}^{8} + x_{2}^{8} + x_{3}^{8} = y_{1}^{8} + y_{2}^{8} + y_{3}^{8}
is not solvable other than trivial permutations of the x_{i} y_{i}. However, it can be shown that there is an infinite number of solutions to,
x_{1}^{8} + x_{2}^{8} + x_{3}^{4} = y_{1}^{8} + y_{2}^{8} + y_{3}^{4}
Proof:
We simply use the identity,
(a+b)^{8} + (ab)^{8} + (4ab)^{4} = 2(a^{4}+14a^{2}b^{2}+b^{4})^{2}
(Note: The RHS is in fact a polynomial invariant of the tetrahedron.) Thus, to solve,
(a+b)^{8} + (ab)^{8} + (4ab)^{4} = (c+d)^{8} + (cd)^{8} + (4cd)^{4}
it suffices to deal with,
a^{4}+14a^{2}b^{2}+b^{4} = c^{4}+14c^{2}d^{2}+d^{4}
Choudhry has given a 7thdeg parametrization (see also Form 3 here) to,
x_{1}^{4}+2nx_{1}^{2}x_{2}^{2}+x_{2}^{4} = y_{1}^{4}+2ny_{1}^{2}y_{2}^{2}+y_{2}^{4}
as,
{x_{1}, x_{2}} = {1+n^{2}t+st^{2}rt^{3}+qt^{4}+3t^{5}+pt^{6}+nt^{7}, n+pt3t^{2}+qt^{3}+rt^{4}+st^{5}n^{2}t^{6}+t^{7}} {y_{1}, y_{2}} = {1n^{2}t+st^{2}+rt^{3}+qt^{4}3t^{5}+pt^{6}nt^{7}, n+pt+3t^{2}+qt^{3}rt^{4}+st^{5}+n^{2}t^{6}+t^{7}}
where {p,q,r,s} = {1n+n^{2}, 2+4n+n^{2}, 4n+n^{3}, 1+n^{2}+n^{3}} for free variable t. For n = 7, and starting with t = 2, we get after removing common factors,
1118^{8} + 1937^{8} + 2502045^{4} = 455^{8} + 1846^{8} + 3200691^{4} and so on for other values of t. 8.1 Eight terms
The first soln to,
x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k }= y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}, k = 8
was found by Nuutti Kuosa in 2006,
1953^{8} + 2012^{8} + 3113^{8} + 861^{8 }= 1128^{8} + 2767^{8 }+ 2557^{8} + 2823^{8}
though it is not known if this the smallest possible as this is the only soln so far. Symmetric linear relations between the terms are, x_{1}x_{2}x_{3}+2x_{4} = (y_{1}y_{2}y_{3}+2y_{4}) 2x_{1}x_{2}x_{3} = y_{1}+y_{2}2y_{3} though there is none between the squares of the terms. It is also not known if k = 1,8 or k = 1,2,8 is possible. Based on known solns to k = 1,3,5, k = 1,2,6, and k = 1,3,7, this author is assuming a k = 1,8 will be found that satisfies the side conditions: x_{1}+x_{2 }= y_{1}+y_{2}, and x_{3}+x_{4 }= y_{3}+y_{4}.
8.2 Nine terms
The first and only soln so far to 8 eighth powers equal to an eighth power,
x_{1}^{8}+x_{2}^{8}+x_{3}^{8}+x_{4}^{8}+x_{5}^{8}+x_{6}^{8}+x_{7}^{8}+x_{8}^{8 } = z^{8}
was found by Scott Chase in 2000,
90^{8} + 223^{8} + 478^{8} + 524^{8} + 748^{8} + 1088^{8} + 1190^{8} + 1324^{8} = 1409^{8}
Whether there are seven 8th powers equal to an 8th power is unknown, though they are conjectured to be possible (the Lander, Parkin, and Selfridge Conjecture in Eulernet). No more k positive kth powers equal to a kth power is known for k > 8 (with the case k = 6 unknown as well), though J. Wroblewski has come close for k = 9,10.
8.3 Ten terms
Other than particular instances of the LetacSinha identity, only two solns are known for multigrade k = 2,4,6,8 (found by Borwein, Lisonek, and Percival in 2000) namely,
[71, 131, 180, 307, 308] = [301, 99, 100, 188, 313] (eq.1) [366, 103, 452, 189, 515] = [508, 245, 18, 331, 471] (eq.2)
which, as signed terms, are also valid for k = 1 hence all known solns are for k = 1,2,4,6,8. (Note: After all, there are 32 possible sums of ± x_{1 }± x_{2 }± x_{3 }± x_{4 }± x_{5 }and an equal number for the other side, and the chances might be that two of the sums will be the same. In general, it raises the question whether systems k = 2,4,…2n as n increases also tend to be valid for k = 1 if the terms are signed.)
Eq.1 has two side conditions: x_{1}x_{2} = y_{1}y_{2} and x_{4}x_{5} = y_{2}y_{3}. Eq. 2 is more structured, satisfying five: x_{1}x_{2} = x_{3}x_{4} = y_{1}y_{2} = y_{2}y_{3}, and x_{1}x_{4} = y_{1}y_{4}, and x_{2}x_{4} = y_{2}y_{4}. It is easily shown that, together with the five eqns k = 1,2,4,6,8, (eq. 2) is the only nontrivial soln of this system of 10 eqns in 10 unknowns. A. Letac, T. Sinha (a+c)^{k} + (ac)^{k} + (3b+d)^{k} + (3bd)^{k} + (4a)^{k} = (3a+c)^{k} + (3ac)^{k} + (b+d)^{k} + (bd)^{k} + (4b)^{k}
for k = 1,2,4,6,8, where a^{2}+12b^{2} = c^{2} (eq.1), and 12a^{2}+b^{2} = d^{2} (eq.2). The ratio a/b = {1/2, 2} must be avoided as it yields trivial solns, with the smallest nontrivial one as {a,b} = {218, 11869}. Labeled as,
x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}+x_{5}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}+y_{5}^{k}
this also satisfies x_{1}x_{2} = y_{1}y_{2}, and x_{3}x_{4} = y_{3}y_{4}, and interestingly, for k = 1,2,
x_{1}^{k}+x_{2}^{k}+x_{5}^{k} = y_{1}^{k}+y_{2}^{k}; x_{3}^{k}+x_{4}^{k} = y_{3}^{k}+y_{4}^{k}+y_{5}^{k},
for any {a,b,c,d}. (In fact, it can be shown that the complete soln to the system with all these constraints is given by this identity.) The first pair of constraints implies that the system k = 2,4,6,8 may have the “side condition”,
x_{1} ± x_{2} = y_{1} ± y_{2}; x_{3} ± x_{4} = y_{3} ± y_{4}
just like for sixth powers k = 2,4,6. To solve the LetacSinha identity, since the complete soln to eq.1 is given by {a,b} = {m^{2}12n^{2}, 2mn}, if we substitute this into eq.2 we get,
4(3m^{4}71m^{2}+432) = d^{2}
where it was set n=1 without loss of generality. Trivial rational solns are given by m = {2, 3, 4, 6}. However, we can use these to generate subsequent ones that are nontrivial, such as m = 109, and so on.
(Update, 2/1/10): Chris Smyth
This is a variation of the LetacSinha 8th power [k.5.5] multigrade. For k = 1,2,4,6,8,
[xy+ax+ayd, (xyaxayd), xybx+by+d, (xy+bxby+d), cx+cy]^{k} = [xy+bx+byd, (xybxbyd), xyax+ay+d, (xy+axay+d), (cxcy)]^{k}
where {a,b,c,d} = {1, 3, 4, 11} and {x,y} satisfies the eqn x^{2}y^{2}13(x^{2}+y^{2})+11^{2 }= 0.
One can see in the general form how “a” in one side has just been replaced with “b” in the other side. With terms expressed as {p_{i},q_{i}}, this obeys,
p_{1}p_{2} = q_{1}q_{2} p_{3}p_{4} = q_{3}q_{4}
If we set that,
p_{1}^{k}+p_{2}^{k}+p_{3}^{k}+p_{4}^{k}+p_{5}^{k} = q_{1}^{k}+q_{2}^{k}+q_{3}^{k}+q_{4}^{k}+q_{5}^{k}, k = 1,2
then one must have 2a^{k}2b^{k}+c^{k} = 0, for k = 1,2, and the consequence is that,
p_{1}^{k}+p_{2}^{k}+p_{5}^{k} = q_{1}^{k}+q_{2}^{k}, p_{3}^{k}+p_{4}^{k} = q_{3}^{k}+q_{4}^{k}+q_{5}^{k}
for k = 1,2 which, naturally enough, is the same set of constraints obeyed by the LetacSinha identity. Expanding the system for k = 4,6,8, it will be seen that the only nontrivial soln in the rationals is {a,b,c,d} = {m, 3m, 4m, 11m^{2}} where one can set m = 1 without loss of generality and {x,y} satisfies the eqn,
x^{2}y^{2}13(x^{2}+y^{2})+11^{2 }= 0
Let {x,y} = {u, v/(u^{2}13)} and this is the elliptic curve,
(u^{2}13)(13u^{2}11^{2}) = v^{2}
Smyth’s form for (k.5.5) can be generalized by adding two terms on each side to create a (k.7.7). A possible addition, given in italics, may be,
[xy+ax+ayc, (xyaxayc), xybx+by+c, (xy+bxby+c), dx+dy, ex+fy, fxey]^{k} = [xy+by+bxc, (xybybxc), xyay+ax+c, (xy+ayax+c), dxdy, ey+fx, exfy]^{k}
which is the same pair successfully added to the (k.4.4). For nonzero {e,f}, whether this has a nontrivial soln up to k = 8, 10, or even 12 is unknown. See also Sixth Powers for the context of this identity. (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 [x_{1}, x_{2}, x_{3}, x_{4}, x_{5}]^{k} = [y_{1}, y_{2}, y_{3}, y_{4}, y_{5}]^{k}, for k = 2,4,6,8,
where x_{1}x_{2} = x_{3}x_{4} = y_{1}y_{2} = y_{2}y_{3} = N, then,
[a+x_{5}, a+y_{4}, a+y_{3}, a+x_{1}, a+x_{3}, a+y_{5}]^{k} = [a+x_{5}, a+y_{4}, a+y_{1}, a+x_{2}, a+x_{4}, a+y_{5}]^{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 et al,
[366, 103, 452, 189, 515] = [508, 245, 18, 331, 471], for k = 2,4,6,8
satisfies this and yields,
[767, 399, 299, 995, 1167, 1205] = [1293, 925, 1279, 57, 115, 679], for k = 1,3,5,7,9
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 also Ninth Powers. (End update.)
(Update, 1/27/10): Piezas
It can be shown that the system,
[x_{1}, x_{2}, x_{3}, x_{4}, x_{5}]^{k} = [y_{1}, y_{2}, y_{3}, y_{4}, y_{5}]^{k}, for k = 2,4,6,8,
with the constraint x_{1}x_{2} = x_{3}x_{4} = y_{1}y_{2} = y_{2}y_{3} can be completely solved in terms of nested square roots. If we add one more constraint to the 8th deg system, namely x_{2}x_{4} = y_{2}y_{4}, we lose one free variable but the problem simplifies to four quadratics to be made squares. This is given by,
[bcd, b+cd, bcd, b+cd, f]^{k} = [a+b2c, a+b, a+b+2c, ab, e]^{k}, for k = 2,4,6,8
where {c,d,e,f} are,
32c^{2} = 5a^{2}10ab+8b^{2} 32d^{2} = 29a^{2}+6ab+72b^{2} 4e^{2} = 9a^{2}6ab+36b^{2} 4f^{2} = 13a^{2}+2ab+4b^{2}
Since the eqns are homogeneous, we can set b = 1 without loss of generality and "a" 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 nontrivial soln is {a,b} = {288, 43} which makes all {c,d,e,f} as squares and yields the original multigrade. 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,
x_{1}^{k} + x_{2}^{k} + x_{3}^{k} + x_{4}^{k} + x_{5}^{k/2} = y_{1}^{k} + y_{2}^{k} + y_{3}^{k} + y_{4}^{k} + y_{5}^{k/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. But 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 nontrivial 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.) Piezas
Theorem: If a^{k}+b^{k}+c^{k}+d^{k}+e^{k} = f^{k}+g^{k}+h^{k}+i^{k}+j^{k}, for k = 2,4,6,8, define n as, a^{4}+b^{4}+c^{4}+d^{4}+e^{4} = n(a^{2}+b^{2}+c^{2}+d^{2}+e^{2})^{2}, then,
72(a^{10}+b^{10}+c^{10}+d^{10}+e^{10}f^{10}g^{10}h^{10}i^{10}j^{10})(a^{14}+b^{14}+c^{14}+d^{14}+e^{14}f^{14}g^{14}h^{14}i^{14}j^{14}) = 35(n+1)(a^{12}+b^{12}+c^{12}+d^{12}+e^{12}f^{12}g^{12}h^{12}i^{12}j^{12})^{2}
which is a part of a family that starts with Ramanujan’s 6108 Identity and depends on a system that is valid for k = 2,4,…2n.
Note 1: It can be shown that the system, (pa+c)^{k} + (pac)^{k} + (qb+d)^{k} + (qbd)^{k} +(ra)^{k} = (qa+c)^{k} + (qac)^{k} + (pb+d)^{k} + (pbd)^{k} +(rb)^{k}, for k = 1,2,4,6,8, where c = √(ma^{2}+nb^{2}) and d = √(na^{2}+mb^{2}) is nontrivial only for {p,q,r} = {1, 3, 4}, and {m,n} = {1, 12}.
Note 2: Just like for k = 1,3,5,7, the general k = 1,2,4,6,8 can be completely parametricized by the form,
(a+bj)^{k }+ (c+dj)^{k }+ (e+fj)^{k }+ (g+hj)^{k }+ (i+j)^{k} = (abj)^{k }+ (cdj)^{k }+ (efj)^{k }+ (ghj)^{k }+ (ij)^{k}
where for simplicity set a = 1 without loss of generality. For the system which also has x_{1}x_{2} = y_{1}y_{2}, x_{3}x_{4} = y_{3}y_{4} it can be shown this entails the necessary but not sufficient condition of finding a nontrivial rational root of a 10^{th} deg resultant. Let d = b, h = f. To satisfy k = 1,2, one can then use the variables i,f. Expanding for k = 4,6,8 gives the eqns,
(Poly01)j^{2} + (Poly02) = 0 (eq.1) (Poly11)j^{4} + (Poly12)j^{2} + (Poly13) = 0 (eq.2) (Poly21)j^{6} + (Poly22)j^{4} + (Poly23)j^{2} + (Poly24) = 0 (eq.3)
Eliminate j between eqs.1,2, then between eq.1,3, (as before, let j = √x for ease of computation) to get two auxiliary resultants, eq.4a and eq.5a, which are 6^{th} and 9^{th} deg in the variable g, respectively. There are still 4 variables, b,c,e,g, and it is at this point that the system for k = 1,2,4,6 with a similar “side condition” has a resultant with a single nontrivial linear factor and some trivial ones. For k = 1,2,4,6,8, one more variable has to be eliminated but since eq.4a and 5a are of high degree, these are horrendous to resolve directly. A trick to reduce the degree (which works for this and similar constraints) is to let {e,g} = {u+v, uv} to get,
(Poly31)v^{4} + (Poly32)v^{2} + (Poly33) = 0 (eq.4b) (Poly41)v^{6} + (Poly42)v^{4} + (Poly43)v^{2} + (Poly44) = 0 (eq.5b)
where again let v = √y to shorten the calculation. Eliminate y and the final resultant turns out to have a nontrivial 10^{th}deg factor in c, with trivial linear ones, showing a qualitative difference from the previous system which had a nontrivial linear factor. One must then find appropriate {b,u} such that this decic has a nontrivial rational root and the LetacSinha identity guarantees there are infinitely many. Whether there is another family such that this decic has rational factors is not known. In summary, the system,
x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}+x_{5}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}+y_{5}^{k}
as S_{7} for k = 1,3,5,7 and S_{8} for k = 2,4,6,8 can have either of the sideconditions, 1) x_{1}x_{2} = x_{3}x_{4} = y_{1}y_{2} 2) x_{1}+x_{2} = x_{3}+x_{4} = y_{1}+y_{2} (or equivalent forms) 3) x_{1}x_{2} = y_{1}y_{2}; x_{3}±x_{4} = y_{3}±y_{4} with S_{7} having solns for all three and S_{8} only for (1), (3). Whether all three can apply to both remains to be seen. Furthermore, for powers greater than the sixth, it seems more constraints are needed to decrease the number of variables so that: a) a parametric soln is possible; or b) the final resultant is not of so high degree.
For example, the LetacSinha identity which is the only one known for eighth powers already satisfies four or five side conditions. This implies that for the 12term nonic and decic systems, and analogous higher ones, a lot of algebraic ingenuity must be employed to find a parametrization. If there is any. Like solving univariate eqns in the radicals, there might come a point when there are simply too many variables than can be handled by the constraints. (See update in Tenth Powers.)
(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+b2c, e]^{k} = [bcd, b+cd, bc+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,
32c^{2} = 5a^{2}10ab+8b^{2} 32d^{2} = 29a^{2}+6ab+72b^{2} 4e^{2} = 9a^{2}6ab+36b^{2} 4f^{2} = 13a^{2}+2ab+4b^{2}
with some {a,b} as trivial like the ratio a/b = {2, 6, etc). A nontrivial soln is {a,b} = {288, 43} which makes all {c,d,e,f} as squares and yields the multigrade 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,
x_{1}^{k} + x_{2}^{k} + x_{3}^{k} + x_{4}^{k} + x_{5}^{k/2} = y_{1}^{k} + y_{2}^{k} + y_{3}^{k} + y_{4}^{k} + y_{5}^{k/2}, for k = 2,4,6,8
(End update.) 8.4 Twelve terms
(Update, 1/17/10): I. x_{1}^{k}+x_{2}^{k} + … + x_{6}^{k} = x_{1}^{k}+x_{2}^{k} + … + x_{6}^{k} , for k = 2,4,8
Wroblewski completely solved the system,
[ap+d, apd, bq+c, bqc, ar, bs]^{k} = [bp+c, bpc, aq+d, aqd, as, br]^{k}, for k = 2,4,8
with the palindromic conditions ma^{2}+nb^{2} = c^{2} and na^{2}+mb^{2} = d^{2} 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, uv, u2v, u+2v; w, v^{2}}
where {u,v,w} satisfy the elliptic curve,
3u^{4}+20u^{2}v^{2}+16v^{4} = w^{2} (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 12a^{2}+b^{2} = c^{2} and a^{2}+12b^{2} = d^{2} 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, 7ad, 5b+c, 5bc, 4a, 8b]^{k} = [7b+c, 7bc, 5a+d, 5ad, 8a, 4b]^{k}
for k = 2,4,8 where 68a^{2}+b^{2} = c^{2} and a^{2}+68b^{2} = d^{2}. And so on for all rational points of eq.1. (End update.)
(Update, 1/12/10): II. x_{1}^{k}+x_{2}^{k} + … + x_{6}^{k} = x_{1}^{k}+x_{2}^{k} + … + x_{6}^{k} , for k = 2,4,6,8
Wroblewski gave three new beautiful identities:
1. If 64a^{2}11b^{2} = 5c^{2} and 11a^{2}+64b^{2} = 5d^{2}, then
(2a+5b+c)^{k} + (2a+5bc)^{k} + (5a2b+d)^{k} + (5a2bd)^{k} + (4a+6b)^{k} + (6a4b)^{k} = (2a5b+c)^{k} + (2a5bc)^{k} + (5a+2b+d)^{k} + (5a+2bd)^{k} + (4a6b)^{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 248a^{2}27b^{2} = 5c^{2} and 27a^{2}+248b^{2} = 5d^{2}, then
(a+10b+c)^{k} + (a+10bc)^{k} + (10ab+d)^{k} + (10abd)^{k} + (a+11b)^{k} + (11ab)^{k} = (a10b+c)^{k} + (a10bc)^{k} + (10a+b+d)^{k} + (10a+bd)^{k} + (a11b)^{k} + (11a+b)^{k}
for k = 2,4,6,8. Trivial ratios are a/b = {1, 3}, while a nontrivial soln is {a,b} = {9533, 3439} from which an infinite more can be computed. These belong to the general form, call this Form 1,
[pa+qb+c, pa+qbc, qapb+d, qapbd, ra+sb, sarb]^{k} = [paqb+c, paqbc, qa+pb+d, qa+pbd, rasb, sa+rb]^{k}
where ma^{2}+nb^{2} = c^{2} and na^{2}+mb^{2} = d^{2} (two quadratics with palindromic coefficients) which also appeared in Tenth Powers by this author. Notice how the variable b is just negated in the RHS. Thus, there are now three known nontrivial solns,
{p,q,r,s} = {1, 3, 2, 8}; {m,n}= {45, 11}; for k = 2,4,6,8,10 {p,q,r,s} = {2, 5, 4, 6}; {m,n}= {64/5, 11/5} {p,q,r,s} = {1, 10, 1, 11}; {m,n}= {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 15a^{2}+96b^{2} = 9c^{2} and 96a^{2}+825b^{2} = 9d^{2}, then
(a+8b+c)^{k} + (a+8bc)^{k} + (3a2b+c)^{k} + (3a2bc)^{k }+ (2ab+d)^{k} + (2abd)^{k} = (a8b+c)^{k} + (a8bc)^{k} + (3a+2b+c)^{k} + (3a+2bc)^{k }+ (2a+b+d)^{k} + (2a+bd)^{k}
for k = 2,4,6,8. Trivial ratios are a/b = {1, 2, 3, 5/2}.
This can be generalized into a Form 2,
[pa+qb+c, pa+qbc, ra+sb+c, ra+sbc, ta+ub+d, ta+ubd]^{k} = [paqb+c, paqbc, rasb+c, rasbc, taub+d, taubd]^{k}
where p_{1}a^{2}+p_{2}b^{2} = c^{2}, and p_{3}a^{2}+p_{4}b^{2} = d^{2}. Again, the variable b has simply been negated in the RHS. Notice the terms obey, x_{1}x_{2} = x_{3}x_{4} = y_{1}y_{2} = y_{3}y_{4} and can be used to form a (k.8.8) for k = 1,3,5,7,9. Whether there are other nontrivial solns of Form 2, or if it can be extended up to k = 10 without being a Form 1, is unknown.
8.5 Fourteen terms
BirckSinha Theorem
Theorem: If a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, k = 2,4, where a+b ≠ c; d+e ≠ f, then,
(a+b+c)^{k} + (a+b+c)^{k} + (ab+c)^{k} + (a+bc)^{k} + (2d)^{k} + (2e)^{k} + (2f)^{k} + 0 = 0 + (2a)^{k} + (2b)^{k} + (2c)^{k} + (d+e+f)^{k} + (de+f)^{k} + (d+ef)^{k} + (d+e+f)^{k}
for k = 1,2,4,6,8.
Note that for any {a,b,c}, then,
0 + (2a)^{k} + (2b)^{k} + (2c)^{k} = (a+b+c)^{k} + (a+b+c)^{k} + (ab+c)^{k} + (a+bc)^{k}, for k = 1,2
and likewise for the {d,e,f}. The zero term is added to show that, if expressed in terms of {x_{i}, y_{i}}, then sums of appropriate terms from opposite sides are equal, namely,
x_{1}+y_{1} = x_{2}+y_{2} = x_{3}+y_{3} = x_{4}+y_{4} = a+b+c x_{5}+y_{5} = x_{6}+y_{6} = x_{7}+y_{7} = x_{8}+y_{8} = d+e+f
which also indicates that it is just a special case of a more general theorem given in the next section.
Proof: (Sinha) Define {s_{2}, s_{4}}:= {a^{2}+b^{2}+c^{2}, a^{4}+b^{4}+c^{4}}, and F_{k}:= (a+b+c)^{k} + (a+b+c)^{k} + (ab+c)^{k} + (a+bc)^{k}  (2a)^{k}  (2b)^{k}  (2c)^{k}
and we find that: F_{1} = 0, F_{2} = 0, F_{4} = 12(s_{2}^{2}2s_{4}), F_{6} = 60s_{2}(s_{2}^{2}2s_{4}), F_{8} = 28(7s_{2}^{2}+2s_{4})(s_{2}^{2}2s_{4})
Given analogous functions t_{1}, t_{2} in the variables {d,e,f}, one can evaluate F_{k}(d,e,f) similarly. Since it is given that a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k} for k = 2,4, then F_{k}(a,b,c) = F_{k}(d,e,f) for k = 1,2,4,6,8. (End proof.) Also, using F_{4} and F_{6}, we get a Ramanujantype identity,
5[a^{2}+b^{2}+c^{2}][(a+b+c)^{4} + (a+b+c)^{4} + (ab+c)^{4} + (a+bc)^{4}  (2a)^{4}  (2b)^{4}  (2c)^{4}] = [(a+b+c)^{6} + (a+b+c)^{6} + (ab+c)^{6} + (a+bc)^{6}  (2a)^{6}  (2b)^{6}  (2c)^{6}]
for arbitrary {a,b,c}. Using a similar analysis, Hirschhorn showed Ramanujan may have used this method to find his 6108 Identity and, in the process, Hirschhorn found an analogous 3710 Identity. We can also use the BirckSinha theorem to find identities with fewer terms. To recall, we have,
(a+b+c)^{k} + (a+b+c)^{k} + (ab+c)^{k} + (a+bc)^{k} + (2d)^{k} + (2e)^{k} + (2f)^{k} + 0 = 0 + (2a)^{k} + (2b)^{k} + (2c)^{k} + (d+e+f)^{k} + (de+f)^{k} + (d+ef)^{k} + (d+e+f)^{k}
If we equate a pair of terms (in color) on one side to the other side,
(a+bc) = (2c) (2f) = (d+ef)
we get, in addition to the requirement a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, for k = {2,4}, the auxiliary conditions that,
a+b = 3c d+e = 3f
Fortunately, such a system is solvable by,
{a,b,c} = {3p+u, 3pu, 2p} {d,e,f} = {3q+v, 3qv, 2q}
(3p+u)^{k} + (3pu)^{k} + (2p)^{k} = (3q+v)^{k} + (3qv)^{k} + (2q)^{k}, k = {2,4}
where,
p^{2}+12q^{2} = u^{2} 12p^{2}+q^{2} = v^{2}
and the BirckSinha theorem becomes the known (k.5.5) BirckSinha Identity discussed previously,
(4p)^{k} + (pu)^{k} + (p+u)^{k} + (2p)^{k} + (3q+v)^{k} + (3qv)^{k} + (2q)^{k} + 0 = 0 + (3p+u)^{k} + (3pu)^{k} + (2p)^{k} + (qv)^{k} + (q+v)^{k} + (2q)^{k} + (4q)^{k}
for k = {1,2,4,6,8} since two terms (as well as the zero term) on either side cancel out. One cannot use the theorem to find another (k.5.5) identity (as trying to cancel other combinations of terms yield trivial results). Wroblewski instead looked for more parametric solutions to,
x_{1}^{k} + x_{2}^{k} + x_{3}^{k} = y_{1}^{k} + y_{2}^{k} + y_{3}^{k}, for k = 2,4
where at least x_{1}+x_{2} = 3x_{3} since this is enough to yield a (k.6.6) identity. Equivalently, let,
{x_{1}, x_{2}, x_{3}} = {3p+u, 3pu, 2p}
and the condition is,
y_{1}^{2} + y_{2}^{2} + y_{3}^{2} = 2(11p^{2}+u^{2}) y_{1}^{4} + y_{2}^{4} + y_{3}^{4} = 2(89p^{4}+54p^{2}u^{2}+u^{4})
though any relation that yields x_{i} = y_{i} must be avoided as it is trivial. There are only seven infinite families known so far. The first is by Birck & Sinha, the second is by Piezas, while the last five are by Wroblewski:
1. y_{1}+y_{2} = 3y_{3}:
(3p+u)^{k} + (3pu)^{k} + (2p)^{k} = (3q+v)^{k} + (3qv)^{k} + (2q)^{k}
p^{2}+12q^{2} = u^{2} 12p^{2}+q^{2} = v^{2}
Ex. {p,q} = {218, 11869}
2. x_{1}+x_{2} = y_{1}+y_{2}:
(3p+u)^{k} + (3pu)^{k} + (2p)^{k} = (3p+v)^{k} + (3pv)^{k} + (2q)^{k}
26p^{2}3q^{2} = u^{2} 24p^{2}q^{2} = v^{2}
Ex. {p,q} = {143, 749}
3. x_{1}x_{2} = y_{1}y_{2}:
(3p+u)^{k} + (3pu)^{k} + (2p)^{k} = (u+v)^{k} + (u+v)^{k} + (2q)^{k}
8p^{2}3q^{2} = 3u^{2} 11p^{2}2q^{2} = v^{2}
Ex. {p,q} = {153, 343}
Note: This yields the nice (k.6.6):
(4p)^{k} + (pu)^{k} + (p+u)^{k} + (u+v)^{k} + (u+v)^{k} + (2q)^{k} = (3p+u)^{k} + (3pu)^{k} + (qu)^{k} + (q+u)^{k} + (q+v)^{k} + (q+v)^{k}, for k = {1,2,4,6,8}
4. 3(x_{1}x_{2}) = 2(y_{1}y_{2}):
(3p+2u)^{k} + (3p2u)^{k} + (2p)^{k} = (3u+v)^{k} + (3u+v)^{k} + (2q)^{k}
8p^{2}3q^{2} = 5u^{2} 3p^{2}+q^{2} = v^{2}
Ex. {p,q} = {323, 397}
OR,
p^{2}q^{2} = 9u^{2} 104p^{2}23q^{2} = 9v^{2}
Ex. {p,q} = {208, 233}
5. x_{1}x_{2} = 3y_{1}2y_{2}:
(3p+u)^{k} + (3pu)^{k} + (2p)^{k} = (2u2v)^{k} + (2u3v)^{k} + (2q)^{k}
33u^{2}134uv+133v^{2} = 153p^{2} 192u^{2}+112uv+937v^{2} = 612q^{2}
Ex. {u,v} = {7,3}
6. 2x_{1}x_{2} = 3(y_{1}y_{2}):
(3p+u)^{k} + (3pu)^{k} + (2p)^{k} = (1/2)^{k }((p+u+v)^{k} + (pu+v)^{k} + (2q)^{k})
19p^{2}+22pu+3u^{2} = 3q^{2} 91p^{2}50pu+3u^{2} = 3v^{2}
Ex. {p,u} = {9, 8}, or {p,u} = {45, 76}
7. 4x_{1}+x_{2} = 3(y_{1}y_{2}):
(3p+u)^{k} + (3pu)^{k} + (2p)^{k} = (1/2)^{k }((5p+u+v)^{k} + (5pu+v)^{k} + (2q)^{k})
67p^{2}+38pu+3u^{2} = 3q^{2} 77p^{2}+106pu3u^{2} = 3v^{2}
Ex. {p,u} = {9, 8}
The two quadratic conditions of each family define an elliptic curve and, from an initial rational point, one can find an infinite more.
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 Multigrade Equations". (End update.)
(Update, 1/17/10): Wroblewski
The BirckSinha Theorem can be expressed in the form,
(p+q)^{k} + (p+q)^{k} + (q+v)^{k} + (q+v)^{k} + (r+u)^{k} + (r+u)^{k} + (2s)^{k} + 0 = (p+v)^{k} + (p+v)^{k} + (2q)^{k} + 0 + (r+s)^{k} + (r+s)^{k} + (s+u)^{k} + (s+u)^{k}
This still has the expected common sums,
x_{1}+y_{1} = x_{2}+y_{2} = x_{3}+y_{3} = x_{4}+y_{4} = q+v x_{5}+y_{5} = x_{6}+y_{6} = x_{7}+y_{7} = x_{8}+y_{8} = s+u
By letting,
{p,q,r,s} = {a+b, c, d+e, f}; {u,v} = {d+e, a+b}
one recovers the BirckSinha Theorem. Wroblewski gave the special case such that the above is valid for k = 1,2,4,6,8 as,
ps = qr p^{2}+3q^{2}+s^{2} = u^{2} q^{2}+r^{2}+3s^{2} = v^{2}
though there are certain {p,q,r,s} such that the system is trivial. Proof: Simply solve for {u,v}, express s in terms of {p,q,r}, and substitute into the system for k = 1,2,4,6,8. Extending to k = 10 will give the trivial values. <End proof>
Assuming r = pt, s = qt, it reduces to two conditions,
p^{2}+3q^{2}+(qt)^{2} = u^{2} q^{2}+(pt)^{2}+3(qt)^{2} = v^{2}
which are two quadratic polynomials to be made squares and can easily be solved as an elliptic curve. An example is {p,q,t} = {9, 8, 2}. Alternatively, by assuming p = 3q, r = 3s, this equates two terms of one side to the other side, {p+q, r+s} = {2q, 2s}. The conditions become,
12q^{2}+s^{2} = u^{2} q^{2}+12s^{2} = v^{2}
Since two terms from each side cancel out, this reduces from a (k.7.7) to the (k.5.5) BirckSinha Identity discussed in a previous section. (End update.)
(Update, 2/2/10): Piezas
More generally, given the same form above,
(p+q)^{k} + (p+q)^{k} + (q+v)^{k} + (q+v)^{k} + (r+u)^{k} + (r+u)^{k} + (2s)^{k} + 0 = (p+v)^{k} + (p+v)^{k} + (2q)^{k} + 0 + (r+s)^{k} + (r+s)^{k} + (s+u)^{k} + (s+u)^{k}
for k = 1,2,4,6,8, then {p,q,r,s,u,v} must satisfy two simultaneous eqns,
p^{2}+2q^{2}+v^{2} = r^{2}+2s^{2}+u^{2} (eq.1) (p^{2}q^{2})(q^{2}v^{2}) = (r^{2}s^{2})(s^{2}u^{2}) (eq.2)
though avoiding trivial cases. Proof: Simply solve for {u,v} in terms of {p,q,r,s}, and substitute into the system. Also, this can be shown as the BirckSinha theorem in disguise by letting {p,q,r,s} = {a+b, c, d+e, f}; {u,v} = {d+e, a+b}, and (eq.1) and (eq.2) are satisfied if,
a^{k}+b^{k}+c^{k }= d^{k}+e^{k}+f^{k}, for k = 2,4
For the special case u = v, then eq.1 becomes,
p^{2}+2q^{2} = r^{2}+2s^{2}
the complete soln of which reduces eq.2 to the easy problem of making a quadratic polynomial into a square. (End update.)
8.6 Sixteen terms
Piezas
Theorem 1. If a^{k}+b^{k}+c^{k}+d^{k} = e^{k}+f^{k}+g^{k}+h^{k}, for k = 2,4 and abcd = efgh, where a+b+c+d ≠ 0 and e+f+g+h ≠ 0, (call the entire system V_{1}) then,
(a+b+c+d)^{k} + (ab+c+d)^{k} + (a+bc+d)^{k} + (a+b+cd)^{k} + (2e)^{k} + (2f)^{k} + (2g)^{k} + (2h)^{k} = (2a)^{k} + (2b)^{k} + (2c)^{k} + (2d)^{k} + (e+f+g+h)^{k} + (ef+g+h)^{k} + (e+fg+h)^{k} + (e+f+gh)^{k}
for k = 1,2,4,6,8. (Call this derived system V_{2}.)
This has the common sums,
x_{1}+y_{1} = x_{2}+y_{2} = x_{3}+y_{3} = x_{4}+y_{4} = a+b+c+d x_{5}+y_{5} = x_{6}+y_{6} = x_{7}+y_{7} = x_{8}+y_{8} = e+f+g+h
Thus if a+b+c+d = e+f+g+h = 0, then x_{i} = y_{i} and the system becomes trivial so this case should be avoided. However, one can set d = h = 0, and this (k.8.8) reduces to the (k.7.7) BirckSinha Theorem.
Proof: Following Sinha’s approach, define,
{s_{1}, s_{2}, s_{4}} = {abcd, a^{2}+b^{2}+c^{2}+d^{2}, a^{4}+b^{4}+c^{4}+d^{4}} and F_{k} = (a+b+cd)^{k} + (a+bc+d)^{k} + (ab+c+d)^{k} + (a+b+c+d)^{k}  (2a)^{k}  (2b)^{k}  (2c)^{k}  (2d)^{k}
and we can evaluate F_{k} for certain k in terms of the s_{i}. Set s_{0} = 8s_{1}+s_{2}^{2}2s_{4}, then,
F_{1} = F_{2} = 0, F_{4} = 12s_{0}, F_{6} = 60s_{0}s_{2}, F_{8} = 28s_{0}(24s_{1}+7s_{2}^{2}+2s_{4}),
For the variables e,f,g,h, assume analogous functions t_{1}, t_{2}, t_{4}, t_{6}. Obviously F_{k}(e,f,g,h) at the same k will be evaluated identically. Since it is given that a^{k}+b^{k}+c^{k}+d^{k} = e^{k}+f^{k}+g^{k}+h^{k}, for k = 2,4, and abcd = efgh, this implies F_{k}(a,b,c,d) = F_{k}(e,f,g,h) for k = 1,2,4,6,8, and we complete the proof. Also, using F_{4} and F_{6}, we can get the more general identity,
5[a^{2}+b^{2}+c^{2}+d^{2}]* [(a+b+cd)^{4} + (a+bc+d)^{4} + (ab+c+d)^{4} + (a+b+c+d)^{4}  (2a)^{4}  (2b)^{4}  (2c)^{4}  (2d)^{4}] = [(a+b+cd)^{6} + (a+bc+d)^{6} + (ab+c+d)^{6} + (a+b+c+d)^{6}  (2a)^{6}  (2b)^{6}  (2c)^{6}  (2d)^{6}]
for arbitrary {a,b,c,d}.
Note: The theorem can be extended to 10th powers if,
a^{k}+b^{k}+c^{k}+d^{k} = e^{k}+f^{k}+g^{k}+h^{k}, for k = 2,4,6
and abcd = efgh, but this has only trivial solns. Proof: Define F_{k} = a^{k}+b^{k}+c^{k}+d^{k}(e^{k}+f^{k}+g^{k}+h^{k}). If F_{k} = 0 for k = 2,4, then,
F_{8} = 4(abcd+efgh)(abcdefgh) + (4/3)(F_{6})(a^{2}+b^{2}+c^{2}+d^{2})
Thus, if both abcdefgh = 0 and F_{6} = 0, this implies F_{8} = 0 as well, in contradiction to Bastien’s Theorem. So, if F_{k} = 0 for k = 2,4,6, then abcd ≠ efgh. Or if abcd = efgh, then F_{k} = 0 only for k = 2,4. <End proof.>
(Update, 2/15/10): Piezas, Wroblewski
Using a numerical search with constraints, Wroblewski found hundreds of solns to a multigrade octic chain of form,
x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}(x_{5}^{k}+x_{6}^{k}+x_{7}^{k}+x_{8}^{k}) = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}(y_{5}^{k}+y_{6}^{k}+y_{7}^{k}+y_{8}^{k}) = z_{1}^{k}+z_{2}^{k}+z_{3}^{k}+z_{4}^{k}(z_{5}^{k}+z_{6}^{k}+z_{7}^{k}+z_{8}^{k})
valid for k = 2,4,6,8. An example is,
[221, 93, 73, 9]^{k}[219, 117, 31, 17]^{k} = [198, 116, 96, 32]^{k}[192, 144, 58, 44]^{k} = [189, 125, 105, 23]^{k}[175, 161, 75, 27]^{k}
Note that it is also true x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k} = x_{5}^{k}+x_{6}^{k}+x_{7}^{k}+x_{8}^{k}, for k = 2,4 and x_{1}x_{2}x_{3}x_{4} = x_{5}x_{6}x_{7}x_{8}, relationships which also hold for the y_{i} and z_{i}. It turns out the general case could be explained in the context of the theorem given by the author in section 8.6 above. Moving half of the terms of one side to the other yields,
(2a)^{k} + (2b)^{k} + (2c)^{k} + (2d)^{k}  ((2e)^{k} + (2f)^{k} + (2g)^{k} + (2h)^{k}) = (a+b+cd)^{k} + (a+bc+d)^{k} + (ab+c+d)^{k} + (a+b+c+d)^{k}  ((e+f+gh)^{k} + (e+fg+h)^{k} + (ef+g+h)^{k} + (e+f+g+h)^{k})
and since either sign of {±d, ±h} can be used and still satisfy abcd = efgh, then the RHS has two distinct octuplets for a given {a,b,c,d,e,f,g,h}. Thus, given,
p_{1}^{k}+p_{2}^{k}+p_{3}^{k}+p_{4}^{k} = q_{1}^{k}+q_{2}^{k}+q_{3}^{k}+q_{4}^{k}, k = 2,4
where p_{1}p_{2}p_{3}p_{4} = q_{1}q_{2}q_{3}q_{4}, then this automatically leads to an octic chain of length 3. Wroblewski’s solns satisfies the constraints,
p_{1}p_{2} = q_{3}q_{4} p_{3}p_{4} = q_{1}q_{2} p_{1}^{2}+p_{2}^{2} = q_{1}^{2}+q_{2}^{2} p_{3}^{2}+p_{4}^{2} = q_{3}^{2}+q_{4}^{2} p_{1}^{4}+p_{2}^{4}+p_{3}^{4}+p_{4}^{4} = q_{1}^{4}+q_{2}^{4}+q_{3}^{4}+q_{4}^{4}
The first author found the complete soln to this system as an elliptic “curve”, namely,
{p_{1}, p_{2}, p_{3}, p_{4}} = {ar+bc, acbr, ds+ef, dfes} {q_{1}, q_{2}, q_{3}, q_{4}} = {arbc, ac+br, dsef, df+es}
r = (d^{2}e^{2})(f/y), s = (a^{2}b^{2})(c/y)
and {a,b,c,d,e,f} must satisfy,
(c^{2}de)(a^{2}b^{2})^{2 }+ (abf^{2})(d^{2}e^{2})^{2} = (abc^{2}+def^{2})y^{2}
which is only a quartic polynomial in {c,f} to be made a square. Given one soln, then an infinite more can be found.
Piezas
Another set of conditions is,
p_{1}p_{4} = q_{1}q_{4} p_{2}p_{3} = q_{2}q_{3} p_{1}+p_{2} = q_{1}+q_{2} p_{1}^{k}+p_{2}^{k}+p_{3}^{k}+p_{4}^{k} = q_{1}^{k}+q_{2}^{k}+q_{3}^{k}+q_{4}^{k}, k = 2,4
The soln can be given as,
{p_{1}, p_{2}, p_{3}, p_{4}} = {a(p+q), b(rs), c(r+s), d(pq)} {q_{1}, q_{2}, q_{3}, q_{4}} = {a(pq), b(r+s), c(rs), d(p+q)}
{p,q,r,s} = {a(b^{2}c^{2}), be, (a^{2}d^{2})b, ae}
and where {a,b,c,d,e} satisfy,
a^{2}(3b^{2}c^{2})+e^{2} = (b^{2}+c^{2})d^{2}
This can be easily solved parametrically as quadratic forms as discussed in Form 17, Fourth Powers . (End update)
(Update, 1/25/10): Wroblewski
Given,
(a^{2})^{k} + (b^{2})^{k} + (a^{2}+b^{2})^{k} = (c^{2})^{k} + (d^{2})^{k} + (c^{2}+d^{2})^{k}, for k = 2,4 (eq.1)
or equivalently,
a^{4}+a^{2}b^{2}+b^{4} = c^{4}+c^{2}d^{2}+d^{4}
then we get the [k.8.8] identity,
[ap+cq, ap+cq, bp+dq, bp+dq, dp+aq, dpaq, cp+bq, cpbq]^{k} = [ap+dq, ap+dq, bp+cq, bp+cq, cp+aq, cpaq, dp+bq, dpbq]^{k}, for k = 1,2,4,8
(Notice how {c,d} of the LHS is replaced by {d,c} in the RHS.) for arbitrary {p,q} and some constants {a,b,c,d}, hence giving an identity in terms of linear forms. An example is {a,b,c,d} = {1, 26, 17, 22}, though Choudhry has given a 7th deg identity for,
a^{4}+ma^{2}b^{2}+b^{4} = c^{4}+mc^{2}d^{2}+d^{4} for arbitrary m. See Form 3 here. To find a [8.7.7], since {p,q} are arbitrary, one can equate appropriate x_{p} = y_{q} so a pair of terms cancel out. Note: Eq. 1, or more generally its unsquared version, is at the core of the Ramanujan 6108 Identity. (End update.)
(Update, 1/17/10): 8.5 Eighteen terms
Lander
(2^{8k+4} + 1)^{8} = (2^{8k+4}1)^{8} + (2^{7k+4})^{8} + (2^{k+1})^{8} + 7(2^{5k+3})^{8} + 7(2^{3k+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. (End update)
(Update, 1/18/10): Wroblewski suggested the family which starts with Pythagorean triples,
(2xy)^{2} + (x^{2}y^{2})^{2} = (x^{2}+y^{2})^{2}
2(4x^{3}y)^{4} + 2(2xy^{3})^{4} + (4x^{4}y^{4})^{4} = (4x^{4}+y^{4})^{4}
(16x^{7}y)^{8} + (2xy^{7})^{8} + 7(8x^{5}y^{3})^{8} + 7(4x^{3}y^{5})^{8} + (16x^{8}y^{8})^{8} = (16x^{8}+y^{8})^{8}
where Lander’s 8th powers identity was just the special case {x,y} = {2^{k}, 1}. If we continue and try to decompose (256x^{16}+1)^{16}  (256x^{16}1)^{16} as a sum of 16th powers, we find the coefficients have prime factors other than 2 and/or 2^{k}1, hence such neat decompositions of nth = 2^{k} powers stop at n = 8. (End update)
