Updates Page 02: Feb

(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 x_i-y_i for all five pairs is divisible by 17.) (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, F_n = 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 of Ninth Powers.

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, 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, 2/15/10): Piezas, Wroblewski

Using a numerical search with constraints, Wroblewski found hundreds of solns to a multi-grade octic chain of form,

x₁^k+x₂^k+x₃^k+x₄^k-(x₅^k+x₆^k+x₇^k+x₈^k) = y₁^k+y₂^k+y₃^k+y₄^k-(y₅^k+y₆^k+y₇^k+y₈^k) = z₁^k+z₂^k+z₃^k+z₄^k-(z₅^k+z₆^k+z₇^k+z₈^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 x₁^k+x₂^k+x₃^k+x₄^k = x₅^k+x₆^k+x₇^k+x₈^k, for k = 2,4 and x₁x₂x₃x₄ = x₅x₆x₇x₈, relationships which also hold for the y_i and z_i. It turns out the general case could be explained in the context of an old result by the first author, namely,

“If aⁿ+bⁿ+cⁿ+dⁿ = eⁿ+fⁿ+gⁿ+hⁿ, for n = 2,4,6 and abcd = efgh, 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.” (If n is only up to n = 4, then k is only up to k = 8.) 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+c-d)^k + (a+b-c+d)^k + (a-b+c+d)^k + (-a+b+c+d)^k - ((e+f+g-h)^k + (e+f-g+h)^k + (e-f+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,

p₁^k+p₂^k+p₃^k+p₄^k = q₁^k+q₂^k+q₃^k+q₄^k, k = 2,4

where p₁p₂p₃p₄ = q₁q₂q₃q₄, then this automatically leads to an octic chain of length 3. Wroblewski’s solns satisfies the constraints,

p₁p₂ = q₃q₄

p₃p₄ = q₁q₂

p₁²+p₂² = q₁²+q₂²

p₃²+p₄² = q₃²+q₄²

p₁⁴+p₂⁴+p₃⁴+p₄⁴ = q₁⁴+q₂⁴+q₃⁴+q₄⁴

The first author found the complete soln to this system as an elliptic “curve”, namely,

{p₁, p₂, p₃, p₄} = {ar+bc, ac-br, ds+ef, df-es}

{q₁, q₂, q₃, q₄} = {ar-bc, ac+br, ds-ef, df+es}

r = (abc²+def²)(d²-e²)(f/y), s = (abc²+def²)(a²-b²)(c/y)

and {a,b,c,d,e,f} must satisfy,

(abc²+def²)(c²de)(a²-b²)² + (abc²+def²)(abf²)(d²-e²)² = y²

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,

p₁p₄ = q₁q₄

p₂p₃ = q₂q₃

p₁+p₂ = q₁+q₂

p₁^k+p₂^k+p₃^k+p₄^k = q₁^k+q₂^k+q₃^k+q₄^k, k = 2,4

The soln can be given as,

{p₁, p₂, p₃, p₄} = {a(p+q), b(r-s), c(r+s), d(p-q)}

{q₁, q₂, q₃, q₄} = {a(p-q), b(r+s), c(r-s), d(p+q)}

{p,q,r,s} = {a(b²-c²), be, (a²-d²)b, ae}

and where {a,b,c,d,e} satisfy,

a²(3b²-c²)+e² = (b²+c²)d²

This can be easily solved parametrically as quadratic forms. (End update)

(Update, 2/9/10): Lee Jacobi, Daniel Madden

The complete soln to a⁴+b⁴ = c⁴+d⁴ and a⁴+b⁴+c⁴ = d⁴ can be reduced to solving an elliptic curve. It turns out that for a special case of four 4th powers equal to a 4th power namely,

a⁴+b⁴+c⁴+d⁴ = (a+b+c+d)⁴ (eq.0)

with variables as ±, then one can do so as well, a small example of which is,

955⁴+1770⁴+(-2634)⁴+5400⁴ = (955+1770-2634+5400)⁴ = 5491⁴

Jacobi's and Madden’s clever soln depended on the identity,

a⁴+b⁴+(a+b)⁴ = 2(a²+ab+b²)²

an identity also useful for solving a⁴+b⁴+c⁴+d⁴+e⁴ = f⁴ as quadratic forms. (See Form 21 here.) Their method starts by adding (a+b)⁴ + (c+d)⁴ to both sides of eq.0,

a⁴+b⁴+(a+b)⁴+c⁴+d⁴+(c+d)⁴ = (a+b)⁴+(c+d)⁴+(a+b+c+d)⁴

and, using the identity, it is seen that eq.0 is equivalent to the special Pythagorean triple,

(a²+ab+b²)² + (c²+cd+d²)² = ((a+b)² + (a+b)(c+d) + (c+d)²)²

Sparing the reader the intermediate algebraic manipulations, using the transformation, {a,b,c,d} = {p+r, p-r, q+s, q-s}, the eqn,

(p+r)⁴+(p-r)⁴+(q+s)⁴+(q-s)⁴ = (2p+2q)⁴

can be solved if {p,q,r,s} satisfy two quadratics to be made squares,

(m²-7)p²+4(m²-1)pq+4(m²-1)q² = (m²+1)r² (eq.1)

8mp²+8mpq-(3m²-8m+3)q² = (m²+1)s² (eq.2)

for some constant m, which is easily proven by solving {r,s} in radicals and substituting it into the eqn to see that it holds. These two quadratic conditions define an elliptic curve. One can then try to find a suitable m and find rational {p,q,r,s}. Or, given a known {p,q,r,s}, a value m can be derived as,

m((p+2q)²-r²) = 3q²+s²

For example, given {a,b,c,d} = {5400, 1770, -2634, 955}, then {p,q,r,s} = {3585, -1679/2, 1815, -3589/2}, and m = 961/61. But this initial soln set can be used to find an infinite more. Since eq.1 and eq.2 are homogeneous, we can assume q = 1 without loss of generality. Starting with a soln to eq.1 as p₁ = -2(3585)/1679, a complete parameterization is,

p = (30/1679) (4291863+112196282u+110805419u²)/(448737-463621u²)

which makes eq.1 a square. But substituting this to eq.2, and still with q = 1, produces a rational quartic polynomial in u that still has to be made a square. To find an initial rational point u, equating p-p₁ = 0 yields the factor 55770003+56098141u = 0, hence this is one suitable value. Treating the quartic polynomial as an elliptic curve, an infinite more can be computed, thus proving that the eqn,

a⁴+b⁴+c⁴+d⁴ = (a+b+c+d)⁴

has an infinite number of distinct, rational solns. (End update.)

(Update, 2/8/10): Piezas

If p⁴+q⁴+r⁴ = 1, then (p+q)⁴ - r⁴ + 1 = 2(p²+pq+q²)² also holds.

(Update, 2/8/10): Demjanenko, Elkies

The complete soln to

p⁴+q⁴+r⁴ = 1 (eq.0)

can be given in the form,

(x+y)⁴ + (x-y)⁴ + z⁴ = 1

where,

ay² = (8mn-3a)x²-2bx-2mn (eq.1)

±az² = 4bx²+8mnx-b (eq.2)

and {a,b} = {2m²+n², 2m²-n²}, for some constants {m,n} hence is a problem of making two quadratics in x as squares and, for appropriate {m,n}, reduces to solving an elliptic curve. One can solve {y,z} in the radicals to see that it holds and it is easily proven that this is the complete soln. Proof: Let {x+y, x-y, z} = {p,q,r}. As eq.1 and eq.2 are also quadratics in {m,n}, solve for m, assume eq.0 as true to rationalize the discriminant, and take the positive sign of the square root to get,

m₁ = n((p+q)²+r²-1) / (2(p²+pq+q²+p+q))

m₂ = n(p²+pq+q²-p-q) / ((p+q)²-r²-1)

One can use the identity given by the author to rationalize the disciminant of m₂. It will then be seen that m₁-m₂ = 0 if eq.0 is true, hence one can always find rational {m,n} in terms of {p,q,r}. For example, using the smallest {p,q,r} = {217519/s, 414560/s, 95800/s} and s = 422481, we get {m,n} = {6007, 30080}. (End proof) However, one can also start with {m,n} as it is possible they are relatively small. Elkies’ first soln used only {m,n} = {8,-5}, and (eq.1) becomes,

153y² = -779x²-206x+80

which has initial soln {x₁,y₁} = {3/14, 1/42}. This yields a parametrization for x as,

x = (51p²-34p-5221)/(238p²+10906)

which, when substituted into (eq.2), becomes the problem of finding a value such that the quartic polynomial in p is a square. This can be treated as an elliptic curve and since x₁ = x = 3/14 has p = -3779/17, from this initial point one can calculate more, proving that eq.0 has an infinite number of distinct solns. Clever transformations may then reduce the size of the coefficients. Explicitly, a particular case can be given as the identity,

(85v²+484v-313)⁴ + (68v²-586v+10)⁴ + (2u)⁴ = (357v²-204v+363)⁴

if u² = 22030+28849v-56158v²+36941v³-31790v⁴,

a soln of which is v = -31/467. This gives, after removing common factors, Elkies' soln. One can then use this initial point to generate an infinite more.

(End update.)

(Update, 2/3/10): Choudhry

Gave a soln to a^k+b^k+c^k = d^k+e^k+f^k, k = 1,2,4, for integer {a,b,c,d,e,f} with the interesting property that one term is equal to 1.

(-p+v)^k + (-q+v)^k + (p+q-2v)^k = (-p-q+2uv)^k + (p+q-2uv-1)^k + 1^k

where

{p,q} = {2u²v+u-v, 4uv+3v+1}

Source: On Representing 1 as the sum or difference of kth powers of integers, The Mathematics Student, Vol. 70, 2001.

Note: Since this also has a+b+c = d+e+f = 0, by the Ramanujan-Hirschhorn Theorem, its terms obey the beautiful relations,

64(a⁶+b⁶+c⁶-d⁶-e⁶-f⁶)(a¹⁰+b¹⁰+c¹⁰-d¹⁰-e¹⁰-f¹⁰) = 45(a⁸+b⁸+c⁸-d⁸-e⁸-f⁸)²

25(a³+b³+c³-d³-e³-f³)(a⁷+b⁷+c⁷-d⁷-e⁷-f⁷) = 21(a⁵+b⁵+c⁵-d⁵-e⁵-f⁵)²

(End update)

(Update, 2/3/10): Choudhry

Sophie Germain’s Identity is given by,

a⁴+4b⁴ = (x²+2xy+2y²)(x²-2xy+2y²)

Choudhry solved the eqn,

a⁴+4b⁴ = c⁴+4d⁴ (eq.1)

using the transformation,

{a,b,c,d} = {(x²+2x-2)z+xy, 2xz+y, (x²+2x+2)z+xy, y}

so that, substituted into eq.1, it had a linear factor in y. This first soln can be used to compute an infinite sequence of polynomial solns, as eq.1 is an elliptic curve in disguise. The transformations, {a,b,c,d} = {p+q, r-s, p-q, r+s}, then {p,q,r,s} = {u, mv, nu, v} reduce eq.1 to the form,

(m-4n³)u²+(m³-4n)v² = 0

so one is to solve,

-(m-4n³)(m³-4n) = z²

One soln is,

{p,q,r,s} = {8x+2x³, -4+2x², 4+4x², -2x+x³}

from which others can then be computed. Source: The Diophantine Equation A⁴+4B⁴ = C⁴+4D⁴, Indian Journal of Pure and Applied Mathematics, Vol. 29, 1998. (End update.)

(Update, 2/3/10): Choudhry

(-8m⁶+2mn⁵)^k + (8m⁵n+n⁶)^k + 2(8m⁶)^k = (8m⁶+2mn⁵)^k + (8m⁵n-n⁶)^k + 2(n⁶)^k, for k = 1,5

A soln in positive terms can be acquired if n > m and within the range 2^2/5 < n/m < 2^3/5, or approx. 1.32 < n/m < 1.51. With terms transposed to one side,

(-8m⁶+2mn⁵)^k + (8m⁵n+n⁶)^k - (8m⁶+2mn⁵)^k - (8m⁵n-n⁶)^k + 2(8m⁶)^k = 2(n⁶)^k

For n = 1, the identity proves that the integer 2 is the sum/difference of six integral 5th powers in infinitely many ways. Source: The Diophantine Equation x₁⁵+x₂⁵+2x₃⁵ = y₁⁵+y₂⁵+2y₃⁵, Ganita, Vol. 48, No. 2, 1997, 115-116.

Note: In one sense, this is a 5th power version of,

(-6x³+y³)³ + (6x³+y³)³ - (6x²y)³ = 2(y³)³

where, for y = 1, proves that 2 infinitely is the sum is the sum/difference of three integral 3rd powers.

Q: Anyone can give a similar identity expressing 2 as integral 7th powers? (End update.)

(Update, 2/2/10): Piezas

More generally, for the equivalent form (disregarding signs),

[a+b, -a+b, b+u, -b+u, c+v, -c+v, 2d]^k = [c+d, -c+d, a+u, -a+u, d+v, -d+v, 2b]^k, for k = 1,2,4,6,8,

the {a,b,c,d,u,v} must satisfy the simultaneous eqns,

a²+2b²+u² = c²+2d²+v²

(a²-b²)(b²-u²) = (c²-d²)(d²-v²)

excluding signed combinations of trivial cases (a+b)(c+d)(b+d)(a+b+c+d)(a²+3b²-c²-3d²) = 0. For the special case u = v, the first condition reduces to,

a²+2b² = c²+2d²

the complete soln of which {a,b,c,d} = {pr+2qs, ps-qr, pr-2qs, ps+qr} reduces the second condition to the easy problem of making a quadratic polynomial into a square. (End update.)

(Update, 2/1/10): This update covers a family that includes 6th, 8th, and 10th power multi-grades. Whether it can be extended for 12th powers with a minimal number of terms remains to be seen.

Piezas (6th powers)

Given the general form,

[xy+ax+3by-c, xy-ax-3by-c, xy-3bx+ay+c, xy+3bx-ay+c]^k =

[xy+ay+3bx-c, xy-ay-3bx-c, xy-3by+ax+c, xy+3by-ax+c]^k

Notice how {x,y} just swap places or alternatively, “a” in one side is replaced by “3b” in the other side. With terms expressed as {p_i, q_i}, this form has the constraints,

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

We can make this valid for k = 1,2,4,6. For k = 4, let c = ab. For k = 6 and some constants {a,b}, then {x,y} must satisfy the simple eqn,

10a²b²-(a²+9b²)(x²+y²)+10x²y² = 0

This is an elliptic “curve” in disguise. Let {x,y} = {u, v/(10u²-a²-9b²)} and one is to solve,

(a²+9b²-10u²)(10a²b²-a²u²-9b²u²) = v²

which is a quartic polynomial in u to be made a square. Trivial ratios are a/b = {0,1,3,9}. It can be proven that for any rational {a,b} that do not have a trivial ratio, one can always find an infinite number of rational {x,y}. Proof: A polynomial soln can be given starting from the trivial point u = 3b. This yields the non-trivial,

u = 9b(a²+15b²)/(5a²-117b²)

from which one can then successively compute an infinite sequence of rational polynomials.

Chris Smyth (8th powers)

This is a variation of the Letac-Sinha 8th power multi-grade. For k = 1,2,4,6,8,

[xy+ax+ay-d, -(xy-ax-ay-d), xy-bx+by+d, -(xy+bx-by+d), cx+cy]^k =

[xy+bx+by-d, -(xy-bx-by-d), xy-ax+ay+d, -(xy+ax-ay+d), cx-cy]^k

where {a,b,c,d} = {1, 3, 4, 11} and {x,y} satisfies the eqn,

x²y²-13(x²+y²)+11² = 0

One can note its affinity with the 6th power multi-grade and see also 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 the constraints,

p₁-p₂ = q₁-q₂

p₃-p₄ = q₃-q₄

If we set that,

p₁^k+p₂^k+p₃^k+p₄^k+p₅^k = q₁^k+q₂^k+q₃^k+q₄^k+q₅^k, k = 1,2

then one must have 2a^k-2b^k+c^k = 0, for k = 1,2, and the consequence is that,

p₁^k+p₂^k+p₅^k = q₁^k+q₂^k,

p₃^k+p₄^k = q₃^k+q₄^k+q₅^k

for k = 1,2 which, naturally enough, is the same set of constraints obeyed by the Letac-Sinha identity. Expanding the system for k = 4,6,8, it will be seen that the only non-trivial soln in the rationals is {a,b,c,d} = {m, 3m, 4m, 11m²} where one can set m = 1 without loss of generality and {x,y} satisfies the eqn,

x²y²-13(x²+y²)+11² = 0

Let {x,y} = {u, v/(u²-13)} and this is the elliptic curve,

(u²-13)(13u²-11²) = v²

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+ay-c, -(xy-ax-ay-c), xy-bx+by+c, -(xy+bx-by+c), dx+dy, ex+fy, fx-ey]^k =

[xy+by+bx-c, -(xy-by-bx-c), xy-ay+ax+c, -(xy+ay-ax+c), dx-dy, ey+fx, ex-fy]^k

which is the same pair successfully added to the (k.4.4) to find the first (k.6.6) identity for 10th powers. For non-zero {e,f}, whether this has a non-trivial soln up to k = 8, 10, or even 12 is unknown.

Choudhry, Wroblewski (10th powers)

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. (End update.)

You can email author at tpiezas@gmail.com.