for k = 1,2,4,6,8, where a²+12b² = c²  (eq.1), and 12a²+b² = d²  (eq.2).  The ratio a/b = {1/2, 2} must be avoided as it yields trivial solns, with the smallest non-trivial one as {a,b} = {218, 11869}.  Labeled as,

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k

this also satisfies,

x₁-x₂ = y₁-y₂;    x₃-x₄ = y₃-y₄

and interestingly enough, for k = 1,2, 

x₁^k+x₂^k+x₅^k = y₁^k+y₂^k;    x₃^k+x₄^k = y₃^k+y₄^k+y₅^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”,

x₁ ± x₂ = y₁ ± y₂;   x₃ ± x₄ = y₃ ± y₄

just like for sixth powers k = 2,4,6.  To solve the Letac-Sinha identity, since the complete soln to eq.1 is given by {a,b} = {m²-12n², 2mn}, if we substitute this into eq.2 we get,

4(3m⁴-71m²+432) = d²

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 non-trivial, such as m = 109, and so on.

Note 1:  It can be shown that the system, (pa+c)^k + (pa-c)^k + (qb+d)^k + (qb-d)^k +(ra)^k = (qa+c)^k + (qa-c)^k + (pb+d)^k + (pb-d)^k +(rb)^k,  for k = 1,2,4,6,8, where c = √(ma²+nb²) and d = √(na²+mb²) is non-trivial 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 = (a-bj)^k + (c-dj)^k + (e-fj)^k + (g-hj)^k + (i-j)^k

where for simplicity set a = 1 without loss of generality.  For the system which also has x₁-x₂ = y₁-y₂,  x₃-x₄ = y₃-y₄ it can be shown this entails the necessary but not sufficient condition of finding a non-trivial 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)j² + (Poly02) = 0                                                         (eq.1)

(Poly11)j⁴ + (Poly12)j² + (Poly13) = 0                                     (eq.2)

(Poly21)j⁶ + (Poly22)j⁴ + (Poly23)j² + (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 non-trivial 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, u-v} to get,

(Poly31)v⁴ + (Poly32)v² + (Poly33) = 0                                    (eq.4b)

(Poly41)v⁶ + (Poly42)v⁴ + (Poly43)v² + (Poly44) = 0               (eq.5b)

where again let v = √y to shorten the calculation.  Eliminate y and the final resultant as a polynomial in c turns out to have a non-trivial 10^th-deg factor (but deg-12 in u and deg-17 in b) with trivial linear ones, showing a qualitative difference from the previous system which had a non-trivial linear factor.  One must then find appropriate b,u such that this decic has a non-trivial rational root and the Letac-Sinha identity guarantees there are infinitely many.  Whether there is another family such that this decic has rational factors is not known.

Note 3:  There are k = 1,3,5,7 with the same constraints as in Note 2 such as,

[37, 3, 53, 19, 51 = [45, 11, 43, 9, 55]

A similar approach can be used for the system k = 1,2,3,5,7 which ends up with a final resultant that is a 5^th-deg in c.  I do not know if this has a parametric soln.  If this system has the related constraint  x₁-x₂ = y₁-y₂,  x₃+x₄ = y₃+y₄, its final resultant is a 6^th-deg in c and also has solns,

[11, 15, 73, -33, -71] = [-57, -53, 1, 39, 65]

[34, -99, 0, -13, 98] = [75, -58, 69, -82, 16]

found by Gloden and Letac in the 1940’s and re-arranged by Shuwen to be valid for k = 2.  The first, based on its other constraints, belong to a family found by this author though the status of the second is still unknown.  (Letac may have a method to find solns such that one term x_i = 0.)  If there is another analysis of the same problems which further reduces the degree, then I would like to know it.  In summary, the system,

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k

as S₇ for k = 1,3,5,7 and S₈ for k = 2,4,6,8 can have either of the side-conditions, 

1)      x₁-x₂ = x₃-x₄ = y₁-y₂

2)      x₁+x₂ = x₃+x₄ = y₁+y₂ (or equivalent forms)

3)      x₁-x₂ = y₁-y₂;  x₃±x₄ = y₃±y₄ 

8.3 Fourteen terms

Birck-Sinha 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 + (a-b+c)^k + (a+b-c)^k + (2d)^k + (2e)^k + (2f)^k =

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

for k = 1,2,4,6,8.  (Note also that (2a)^k + (2b)^k + (2c)^k = (a+b+c)^k + (-a+b+c)^k + (a-b+c)^k + (a+b-c)^k,  for k = 1,2.)

Proof: (Sinha) Define {s₂, s₄}:= {a²+b²+c²,  a⁴+b⁴+c⁴}, and F_k:= (a+b+c)^k + (-a+b+c)^k + (a-b+c)^k + (a+b-c)^k - (2a)^k - (2b)^k - (2c)^k 

and we find that:  F₁ = 0,    F₂ = 0,   F₄ = 12(s₂²-2s₄),   F₆ = 60s₂(s₂²-2s₄),    F₈ = 28(7s₂²+2s₄)(s₂²-2s₄)

Given analogous functions t₁, t₂ 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.)  An extension to tenth powers can also be given . Also, using F₄ and F₆, we get a Ramanujan-type identity,

for arbitrary {a,b,c}.  Using a similar analysis, Hirschhorn showed Ramanujan may have used this method to find his 6-10-8 Identity and, in the process, Hirschhorn found an analogous 3-7-10 Identity.

8.4  Sixteen terms

Piezas

Theorem. If a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k,  for k = 2,4,6, where a+b ≠ ±(c+d); e+f ≠ ±(g+h), and abcd = efgh, (call the entire system V₁) then,

(a+b+c-d)^k + (a+b-c+d)^k + (a-b+c+d)^k + (-a+b+c+d)^k + (2e)^k + (2f)^k + (2g)^k + (2h)^k =

(e+f+g-h)^k  + (e+f-g+h)^k  + (e-f+g+h)^k  + (-e+f+g+h)^k  + (2a)^k + (2b)^k + (2c)^k + (2d)^k

for k = 1,2,4,6,8,10.  (Call this derived system V₂.)

Proof:  Following Sinha’s approach, define,

{s₁, s₂, s₄, s₆} = {abcd, a²+b²+c²+d²,  a⁴+b⁴+c⁴+d⁴,  a⁶+b⁶+c⁶+d⁶} and F_k = (a+b+c-d)^k + (a+b-c+d)^k + (a-b+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 s₀ = -8s₁+s₂²-2s₄, then,

F₁ = F₂ = 0,    F₄ = 12s₀,    F₆ = 60s₀s₂,    F₈ = 28s₀(-24s₁+7s₂²+2s₄),   F₁₀ = 20s₀(-264s₁s₂+37s₂³-18s₂s₄+32s₆)

For the variables e,f,g,h, assume analogous functions t₁, t₂, t₄, t₆.  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 = 2,4,6, and abcd = efgh, this implies F_k(a,b,c,d) = F_k(e,f,g,h) for k = 1,2,4,6,8,10 and we complete the proof.  Also, using F₄ and F₆, we can get the more general identity,

5[a²+b²+c²+d²]*

[(a+b+c-d)⁴ + (a+b-c+d)⁴ + (a-b+c+d)⁴ + (-a+b+c+d)⁴ - (2a)⁴ - (2b)⁴ - (2c)⁴ - (2d)⁴] =

[(a+b+c-d)⁶ + (a+b-c+d)⁶ + (a-b+c+d)⁶ + (-a+b+c+d)⁶ - (2a)⁶ - (2b)⁶ - (2c)⁶ - (2d)⁶]

for arbitrary {a,b,c,d}.  Note:  It seems Shuwen’s work (et al) imply that the system,

aⁿ+bⁿ+cⁿ+dⁿ = eⁿ+fⁿ+gⁿ+hⁿ, for n = 1,2,3, and abcd = efgh

of which V₁ reduces to after taking square roots has only trivial solns.  For the simpler case when V₁ is only up to k = 2,4, there are parametric solns as given by this author in Fourth Powers but its derived system V₂ is valid only up to k = 8.  (If in addition, d=h=0, then the requirement abcd = efgh disappears and we get the Birck-Sinha theorem.)

PART 12.  Sum / Sums of Ninth Powers and Higher 

I. Ninth Powers

Results for these came only in the last decade and are only numerical.  For ten terms, or an equal sum of five ninth powers,

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k   (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}

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k+x₆^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k+y₆^k

the multi-grade system k = 1,3,5,7,9 has only two known solns.  The first is, 

[7, 91, 173, 269, 289, 323] = [29, 59, 193, 247, 311, 313]

found by Shuwen in 2000 after two months of computer time!  If the special case x₁ = 0 is found, this will lead to an ideal soln of deg 10 of the Prouhet-Tarry-Escott problem.  This soln has two special properties.  First, analogous to the situation for even systems k = 2,4,…2n which can be made valid for odd k = 1, it seems odd systems k = 1,3,…2n+1 can be made true for even k = 2 since one can “move” terms around and there might be a balanced partition,

x₁^k+x₂^k+…+x_m^k = y₁^k+y₂^k+…+y_m^k

also true for k = 2.  And in fact there is:  [-269, -173, -7, 29, 311, 313] = [-247, -193, -59, 91, 289, 323].  The consequence of letting a system valid for an even power is that it “fixes” the position of the terms and one can no longer transpose them arbitrarily.  Furthermore, two remarks:  First, recall that we extended a result for fifth powers and gave a conjecture for the seventh as,

"Let F_k:= x₁^k+x₂^k+x₃^k+x₄^k+x₅^k – (y₁^k+y₂^k+y₃^k+y₄^k+y₅^k).  If F_k = 0 for k = 1,2,3,5,7, then (11F₉)(x₁²+x₂²+x₃²+x₄²+x₅²) = 9F₁₁."

Its analogue for ninth powers is,

"Let F_k:= x₁^k+x₂^k+x₃^k+x₄^k+x₅^k+x₆^k – (y₁^k+y₂^k+y₃^k+y₄^k+y₅^k+y₆^k).  If F_k = 0 for k = 1,2,3,5,7,9, then (13F₁₁)(x₁²+x₂²+x₃²+x₄²+x₅²+x₆²) = 11F₁₃.

It is readily verified that Shuwen's soln satisfies this.  The pattern is quite obvious and can be extrapolated for systems with powers 11^th, 13^th, 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+…+x_m = y₁+y₂+…+y_m = 0.  It turns out that for this particular example one can do so in four ways, namely,

[7, 173, 289, -247, 91, -313] = [-269, -323, 59, 311, 29, 193]

[7, 173, 289, -247, -29, -193] = [-269, -323, 59, 311, -91, 313]

since 91-313 = -29-193, and,

[7, 289, -59, -313, 269, -193] = [-91, -173, 29, 311, -323, 247]

[7, 289, -59, -313, 323, -247] = [-91, -173, 29, 311, -269, 193]

since 269-193 = 323-247.  Of course, these versions are still valid for k = 1,3,5,7,9.

Update, 11/27/09:  Jarek Wroblewski has found a second soln to (k.6.6) for k = 1,3,5,7,9.  This is,

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

Just like Shuwen's, there is a 6.6 partition such that it is for k = 2 as well,

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

This form satisfies the conjecture involving 11th and 13th powers give above.  It can also be re-arranged so that sums of each side are equal to zero,

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

since 57+995 = -115+1167.  All these versions are still valid for k = 1,3,5,7,9.  (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,

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

for k = 1,2,3,9 have 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 "side conditions":

x₁-y₁ = x₂-y₂ = x₃-y₃ = -(x₄-y₄) = -(x₅-y₅) = -(x₆-y₆)

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 u_i, v_i 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

Define,

{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 x_i, y_i.  This author started with those and derived the u_i, v_i 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 u_i are just negated in the v_i.  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 x_i, y_i as,

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

y_i = {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 x_i  are just negated in the y_i.)  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 x_i, y_i in the first method.  In 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.)

II. Tenth Powers

For the even multi-grade system k = 2,4,6,8,10, there is a different curious result.  The only known soln (pls see update below) is,

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

found by Kuosa, Meyrignac (in 1999) 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,

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 = 161

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 = 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

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

not counting seven other combinations which simply negate the terms.  Whether this is: (1) a fluke, or (2) a property of the 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.  The second combination (arranged differently) also satisfies x₁+x₂ = y₁+y₂; x₃+x₄+x₅+x₆ = y₃+y₄+y₅+y₆ though again it is unknown if two or all three,

x₁ ± x₂ = y₁ ± y₂

x₃ ± x₄ = y₃ ± y₄

x₅ ± x₆ = y₅ ± y₆

can be satisfied by another soln to this decic system.  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):  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.  Choudhry and Wroblewski  (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) 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.

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.