028: Eighth Powers

and so on for other values of t.

8.1 Eight terms

The first soln to,

x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^k, k = 8

was found by Nuutti Kuosa in 2006,

1953⁸ + 2012⁸ + 3113⁸ + 861⁸ = 1128⁸ + 2767⁸ + 2557⁸ + 2823⁸

1118⁸ + 1937⁸ + 2502045⁴ = 455⁸ + 1846⁸ + 3200691⁴

(Update, 1/18/10): Wroblewski suggested the family which starts with Pythagorean triples,

(2xy)² + (x²-y²)² = (x²+y²)²

2(4x³y)⁴ + 2(2xy³)⁴ + (4x⁴-y⁴)⁴ = (4x⁴+y⁴)⁴

(16x⁷y)⁸ + (2xy⁷)⁸ + 7(8x⁵y³)⁸ + 7(4x³y⁵)⁸ + (16x⁸-y⁸)⁸ = (16x⁸+y⁸)⁸

where Lander’s 8th powers identity was just the special case {x,y} = {2^k, 1}. If we continue and try to decompose (256x¹⁶+1)¹⁶ - (256x¹⁶-1)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)

(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 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+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 = (d²-e²)(f/y), s = (a²-b²)(c/y)

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

(c²de)(a²-b²)² + (abf²)(d²-e²)² = (abc²+def²)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 as discussed in Form 17, Fourth Powers . (End update)

(Update, 1/25/10): Wroblewski

Given,

(a²)^k + (b²)^k + (a²+b²)^k = (c²)^k + (d²)^k + (c²+d²)^k, for k = 2,4 (eq.1)

or equivalently,

a⁴+a²b²+b⁴ = c⁴+c²d²+d⁴

then we get the [k.8.8] identity,

[ap+cq, -ap+cq, bp+dq, -bp+dq, dp+aq, dp-aq, cp+bq, cp-bq]^k =

[ap+dq, -ap+dq, bp+cq, -bp+cq, cp+aq, cp-aq, dp+bq, dp-bq]^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,

a⁴+ma²b²+b⁴ = c⁴+mc²d²+d⁴

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 6-10-8 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)

8.4 Twelve terms

(Update, 1/17/10): I. x₁^k+x₂^k + … + x₆^k = x₁^k+x₂^k + … + x₆^k , 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.)

(Update, 1/12/10): II. x₁^k+x₂^k + … + x₆^k = x₁^k+x₂^k + … + x₆^k , for k = 2,4,6,8

Wroblewski gave three 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, call this Form 1,

[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 (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 non-trivial 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 -15a²+96b² = 9c² and -96a²+825b² = 9d², then

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

(a-8b+c)^k + (a-8b-c)^k + (3a+2b+c)^k + (3a+2b-c)^k + (2a+b+d)^k + (2a+b-d)^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+qb-c, ra+sb+c, ra+sb-c, ta+ub+d, ta+ub-d]^k =

[pa-qb+c, pa-qb-c, ra-sb+c, ra-sb-c, ta-ub+d, ta-ub-d]^k

where p₁a²+p₂b² = c², and p₃a²+p₄b² = d². Again, the variable b has simply been negated in the RHS. Notice the terms obey,

x₁-x₂ = x₃-x₄ = y₁-y₂ = y₃-y₄

and can be used to form a (k.8.8) for k = 1,3,5,7,9. Whether there are other non-trivial 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

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 + 0 =

0 + (2a)^k + (2b)^k + (2c)^k + (-d+e+f)^k + (d-e+f)^k + (d+e-f)^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 + (a-b+c)^k + (a+b-c)^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,

x₁+y₁ = x₂+y₂ = x₃+y₃ = x₄+y₄ = a+b+c

x₅+y₅ = x₆+y₆ = x₇+y₇ = x₈+y₈ = 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 {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.) Also, using F₄ and F₆, we get a Ramanujan-type identity,

5[a²+b²+c²][(a+b+c)⁴ + (-a+b+c)⁴ + (a-b+c)⁴ + (a+b-c)⁴ - (2a)⁴ - (2b)⁴ - (2c)⁴] =

[(a+b+c)⁶ + (-a+b+c)⁶ + (a-b+c)⁶ + (a+b-c)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 6-10-8 Identity and, in the process, Hirschhorn found an analogous 3-7-10 Identity. We can also use the Birck-Sinha theorem to find identities with fewer terms. To recall, we have,

(a+b+c)^k + (-a+b+c)^k + (a-b+c)^k + (a+b-c)^k + (2d)^k + (2e)^k + (2f)^k + 0 =

0 + (2a)^k + (2b)^k + (2c)^k + (-d+e+f)^k + (d-e+f)^k + (d+e-f)^k + (d+e+f)^k

If we equate a pair of terms (in color) on one side to the other side,

(a+b-c) = (2c)

(2f) = (d+e-f)

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, 3p-u, 2p}

{d,e,f} = {3q+v, 3q-v, 2q}

(3p+u)^k + (3p-u)^k + (2p)^k = (3q+v)^k + (3q-v)^k + (2q)^k, k = {2,4}

where,

p²+12q² = u²

12p²+q² = v²

and the Birck-Sinha theorem becomes the known (k.5.5) Birck-Sinha Identity discussed previously,

(4p)^k + (p-u)^k + (p+u)^k + (2p)^k + (3q+v)^k + (3q-v)^k + (2q)^k + 0 =

0 + (3p+u)^k + (3p-u)^k + (2p)^k + (q-v)^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,

x₁^k + x₂^k + x₃^k = y₁^k + y₂^k + y₃^k, for k = 2,4

where at least x₁+x₂ = 3x₃ since this is enough to yield a (k.6.6) identity. Equivalently, let,

{x₁, x₂, x₃} = {3p+u, 3p-u, 2p}

and the condition is,

y₁² + y₂² + y₃² = 2(11p²+u²)

y₁⁴ + y₂⁴ + y₃⁴ = 2(89p⁴+54p²u²+u⁴)

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. y₁+y₂ = 3y₃:

(3p+u)^k + (3p-u)^k + (2p)^k = (3q+v)^k + (3q-v)^k + (2q)^k

p²+12q² = u²

12p²+q² = v²

Ex. {p,q} = {218, 11869}

2. x₁+x₂ = y₁+y₂:

(3p+u)^k + (3p-u)^k + (2p)^k = (3p+v)^k + (3p-v)^k + (2q)^k

26p²-3q² = -u²

24p²-q² = -v²

Ex. {p,q} = {143, 749}

3. x₁-x₂ = y₁-y₂:

(3p+u)^k + (3p-u)^k + (2p)^k = (u+v)^k + (-u+v)^k + (2q)^k

8p²-3q² = -3u²

11p²-2q² = v²

Ex. {p,q} = {153, 343}

Note: This yields the nice (k.6.6):

(4p)^k + (p-u)^k + (p+u)^k + (u+v)^k + (-u+v)^k + (2q)^k =

(3p+u)^k + (3p-u)^k + (q-u)^k + (q+u)^k + (-q+v)^k + (q+v)^k, for k = {1,2,4,6,8}

4. 3(x₁-x₂) = 2(y₁-y₂):

(3p+2u)^k + (3p-2u)^k + (2p)^k = (3u+v)^k + (-3u+v)^k + (2q)^k

8p²-3q² = 5u²

3p²+q² = v²

Ex. {p,q} = {323, 397}

OR,

p²-q² = -9u²

104p²-23q² = 9v²

Ex. {p,q} = {208, 233}

5. x₁-x₂ = 3y₁-2y₂:

(3p+u)^k + (3p-u)^k + (2p)^k = (2u-2v)^k + (2u-3v)^k + (2q)^k

33u²-134uv+133v² = 153p²

-192u²+112uv+937v² = 612q²

Ex. {u,v} = {7,3}

6. 2x₁-x₂ = 3(y₁-y₂):

(3p+u)^k + (3p-u)^k + (2p)^k = (1/2)^k ((p+u+v)^k + (-p-u+v)^k + (2q)^k)

19p²+22pu+3u² = 3q²

91p²-50pu+3u² = 3v²

Ex. {p,u} = {9, -8}, or {p,u} = {45, 76}

7. 4x₁+x₂ = 3(y₁-y₂):

(3p+u)^k + (3p-u)^k + (2p)^k = (1/2)^k ((5p+u+v)^k + (-5p-u+v)^k + (2q)^k)

67p²+38pu+3u² = 3q²

77p²+106pu-3u² = -3v²

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 Multi-grade Equations". (End update.)

(Update, 1/17/10): Wroblewski

The Birck-Sinha 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,

x₁+y₁ = x₂+y₂ = x₃+y₃ = x₄+y₄ = q+v

x₅+y₅ = x₆+y₆ = x₇+y₇ = x₈+y₈ = s+u

By letting,

{p,q,r,s} = {-a+b, c, -d+e, f}; {u,v} = {d+e, a+b}

one recovers the Birck-Sinha Theorem. Wroblewski gave the special case such that the above is valid for k = 1,2,4,6,8 as,

ps = qr

p²+3q²+s² = u²

q²+r²+3s² = v²

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,

p²+3q²+(qt)² = u²

q²+(pt)²+3(qt)² = v²

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,

12q²+s² = u²

q²+12s² = v²

Since two terms from each side cancel out, this reduces from a (k.7.7) to the (k.5.5) Birck-Sinha 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,

p²+2q²+v² = r²+2s²+u² (eq.1)

(p²-q²)(q²-v²) = (r²-s²)(s²-u²) (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 Birck-Sinha 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,

p²+2q² = r²+2s²

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 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 =

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

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

This has the common sums,

x₁+y₁ = x₂+y₂ = x₃+y₃ = x₄+y₄ = a+b+c+d

x₅+y₅ = x₆+y₆ = x₇+y₇ = x₈+y₈ = 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) Birck-Sinha Theorem.

Proof: Following Sinha’s approach, define,

{s₁, s₂, s₄} = {abcd, a²+b²+c²+d², a⁴+b⁴+c⁴+d⁴} 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),

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, 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 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: 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,

F₈ = -4(abcd+efgh)(abcd-efgh) + (4/3)(F₆)(a²+b²+c²+d²)

Thus, if both abcd-efgh = 0 and F₆ = 0, this implies F₈ = 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.>