024: Sixth Powers (6 or 8 terms)

Theorem: If (a-b-c)^k + (a-b+c)^k + (b+3c)^k + (b+c)^k = (a+b-d)^k + (a+b+d)^k + (-b+3c)^k + (-b+c)^k, for k = 1,3,5, then,

(a+b+c+d)^k + (a+b+c-d)^k + (-a+b+2c)^k + (b-4c)^k = (a+b-c+d)^k + (a+b-c-d)^k + (-a+b-2c)^k + (b+4c)^k, for k = 1,2,4,6,

where {a,b,c,d} satisfy a²+3ab+4b² = 8c² and a²-5ab+36b² = 8d². Note how in the second system, the LHS difers from the RHS only in the sign of the variable c. This identity is easily proven by solving for {c,d} in terms of {a,b} and subsituting into the two systems, though the ratio a/b = {-4, -2, -1, 1, 4} should be avoided as they are trivial, with a small non-trivial soln as {a,b} = {14,1}. (End note.)

6.4 Eight terms

Some general identities analogous to the quadratic-quartic system Q₁ are:

Theorem (Piezas): Let a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k, for k = 2,4,6.

Define {x,y,z} = {a²+b²+c²+d², a⁴+b⁴+c⁴+d⁴, a⁶+b⁶+c⁶+d⁶}. Then,

a⁸+b⁸+c⁸+d⁸-e⁸-f⁸-g⁸-h⁸ = -4(a²-e²)(b²-e²)(c²-e²)(d²-e²) = -4(abcd)²+4(efgh)²

4(a¹⁰+b¹⁰+c¹⁰+d¹⁰-e¹⁰-f¹⁰-g¹⁰-h¹⁰) = 5(a⁸+b⁸+c⁸+d⁸-e⁸-f⁸-g⁸-h⁸)(x)

8(a¹²+b¹²+c¹²+d¹²-e¹²-f¹²-g¹²-h¹²) = 6(a⁸+b⁸+c⁸+d⁸-e⁸-f⁸-g⁸-h⁸)(x²+y)

24(a¹⁴+b¹⁴+c¹⁴+d¹⁴-e¹⁴-f¹⁴-g¹⁴-h¹⁴) = 7(a⁸+b⁸+c⁸+d⁸-e⁸-f⁸-g⁸-h⁸)(x³+3xy+2z)

The first one is true regardless of whether e is replaced by any of {f,g,h}. The next two can be combined together to an 8-12-10 Identity. Define n as,

a⁴+b⁴+c⁴+d⁴ = n(a²+b²+c²+d²)²

then,

25(a⁸+b⁸+c⁸+d⁸-e⁸-f⁸-g⁸-h⁸)(a¹²+b¹²+c¹²+d¹²-e¹²-f¹²-g¹²-h¹²) = 12(n+1)(a¹⁰+b¹⁰+c¹⁰+d¹⁰-e¹⁰-f¹⁰-g¹⁰-h¹⁰)²

This is analogous to Ramanujan’s 6-10-8 Identity though since Ramanujan was able to use a linear relationship between the terms a_i, namely a+b = ±c, to define n as the constant ½, his result was then much neater. However, work by Euler shows that,

a⁴+b⁴+c⁴+d⁴ = n(a²+b²+c²+d²)²

does have a parametric soln for n = ½, given in the section on Fourth Powers, though it is much more complicated than for the case d = 0.

Chernick (and Escott, independently)

There is the aesthetically-pleasing identity,

(-u+7w)^k + (u-2v+w)^k + (-3u-w)^k + (3u+2v+w)^k =

(u+7w)^k + (-u+2v+w)^k + (3u-w)^k + (-3u-2v+w)^k

for k = 1,2,4,6, if u²+uv+v² = 7w²

with signs changed by this author so it would be valid for k = 1. This was derived using the equally simple quartic identity,

u^k + v^k + (u+v)^k = w^k + (2w)^k + (3w)^k

for k = 2,4 with the same condition. Since the eqn is homogeneous, one can do the substitution {u,v,w} = {y+z, y-z, x} and the condition becomes,

7x²-3y² = z²

where the binary quadratic form has square-free discriminant D = 21. This is the only known quadratic identity (in contrast to the system k = 1,3,5,7 which has three known D). So how did Chernick and Escott find this? One technique is to keep applying the transformation theorems discussed in Chap II to gradually increase the range of the exponents and, if the chosen starting identity has special properties, the same terms may appear on both sides and cancel out, keeping down the number of terms. For example, using the quartic identity and Theorem 5 with T = 0, this becomes,

[u, v, u+v, -u, -v, -u-v] = [w, 2w, 3w, -w, -2w, -3w], k = 1,2,..5

If the Tarry-Escott theorem is then used, generally this doubles the number of terms. But for this particular instance, using T = w, the number of terms on either side of the equation instead of doubling, increases only up to eight but the range is now k = 1,2,3,4,5,6. A second application of the same theorem, with T = u, does not increase the number of terms at all and produces a symmetric ideal soln for k = 1,2,..7. Since this has appropriate pairs with a common sum, using Theorem 5 in reverse it is easy to recover a (k.4.4) for k = 2,4,6 and which is the identity given at the start of this section.

Using a similar method, Tarry earlier found the first ideal soln (symmetric) for k =1,2,..7. Starting with [1, 9] = [4, 6] which is only for k = 1, he used five successive applications of the Tarry-Escott theorem with T = 2, 1, 7, 8, 13, respectively, to get,

[1, 5, 10, 16, 27, 28, 38, 39] = [2, 3, 13, 14, 25, 31, 36, 40], k = 1,2,…6

Note that this is a symmetric soln with the common sum 41, implying this is dependent on an identity valid for k = 1,3,5. Then using a last transformation with T = 11 found the ideal soln,

[1, 5, 10, 24, 28, 42, 47, 51] = [2, 3, 12, 21, 31, 40, 49, 50], k = 1,2,…7

It turns out this can be generalized with solns depending on the nice eqn x²+7y² = 8z².

Piezas

To simplify matters we will start with the sixth deg. Rearranging terms, this is

[5, 16, 27, 38, 28, 39, 1, 10] = [3, 14, 25, 36, 2, 13, 40, 31], k = 1,2,…6

which makes it easy to see that the first six terms (with a skip after the fourth) of each side involve an arithmetic progression, differing by 11. This is then an example of a “special” property. To recall, by Frolov’s theorem we can always set a₁ = 0 so, to generalize Tarry's result, we are to solve the system,

[0, a, 2a, 3a, a+c, 2a+c, x₁, x₂] = [b, a+b, 2a+b, 3a+b, a+b-c, 2a+b-c, y₁, y₂], k = 1,2,…6

with unknowns {a,b,c,x₁,x₂,y₁,y₂}. Using the Tarry-Escott theorem, with T = a the properties of the system ensure there is no increase in the number of terms and this becomes,

[0, a+c, x₁, x₂, 4a+b, 3a+b-c, a+y₁, a+y₂] = [b, a+b-c, y₁, y₂, 4a, 3a+c, a+x₁, a+x₂]

now for k = 1,2,…7. But since we wish this to be symmetric, pairs must have a common sum,

0 + (4a+b) = (a+c) + (3a+b-c) = (x₁) + (a+y₁) = (x₂) + (a+y₂)

b + (4a) = (a+b-c) + (3a+c) = (y₁) + (a+x₁) = (y₂) + (a+x₂)

where the first two pairs have the sum 4a+b while the last two imply x₁+y₁ = x₂+y₂. (In Tarry’s example, this is 1+40 = 10+31.) This can be satisfied by letting {x₁,x₂,y₁,y₂} = {t+u, t+v, t-u, t-v} and, if these are to have the same common sum as the others, then it must be the case that,

(t+u) + (a+(t-u)) = 4a+b

so t = (3a+b)/2. We have reduced our unknowns to just five: {a,b,c,u,v}. (In fact, it can be assumed c = 1 without loss of generality.) Using Theorem 5 in reverse, if we subtract (4a+b)/2 from the first four terms on each side, namely {0, a+c, x₁, x₂} and {b, a+b-c, y₁, y₂}, this gives us,

(4a-b)^k + (2a-b+2c)^k + (a+2u)^k + (a+2v)^k = (4a+b)^k + (2a+b-2c)^k + (a-2u)^k + (a-2v)^k

for k = 2,4,6. To simplify, let {a,b,c} = {2p, -2q, -q+r} and rearranging terms, we get,

(p+u)^k + (p+v)^k + (2p+r)^k + (4p+q)^k = (p-u)^k + (p-v)^k + (2p-r)^k + (4p-q)^k, (eq.1)

Using this simpler system, we can then solve for the unknowns. This is given by,

{p,q,r,u,v} = {2x²+x+1, x²+6x+1, -8(x²+2x), 6x²+4x-2+y, 6x²+4x-2-y}

where x,y must satisfy,

y² = 15x⁴+104x³+223x²+54x+4

with integral solns x = -3,-2,0,1,5 which are trivial though a non-trivial one is x = -1/7. From these one can then compute more rational points. This was derived by expanding eq.1 at k = 2,4,6 and eliminating {u,v} to get the first condition 8p²-8q²-5qr-r² = 0. Letting {p,q,r} = {z, 2y, x-5y}, this becomes,

x²+7y² = 8z²

with initial soln {x,y,z}= {1,1,1} and is easily solved parametrically. By recovering {u,v} there is a second quadratic condition to fulfill hence the elliptic curve. As was alluded to earlier, this symmetric sixth degree implied a fifth degree identity. Thus, in summary, if,

(3p+q)^k + (p+q)^k + (p+r)^k + (u)^k = (3p-q)^k + (p-q)^k + (p-r)^k + (-v)^k, for k = 1,3,5, then,

(p-u)^k + (p-v)^k + (2p-r)^k + (-4p+q)^k = (-p-u)^k + (-p-v)^k + (-2p-r)^k + (4p+q)^k, for k = 1,2,4,6.

Note that the second system satisfies x₁-y₁ = x₂-y₂ = (x₃-y₃)/2 = -(x₄-y₄)/4 and a minor change in signs made it valid for k = 1 as well. From this one can find an ideal seventh degree soln such as the beautifully concise,

q^k + (2p+r)^k + (3p+u)^k + (3p+v)^k + (5p-u)^k + (5p-v)^k + (6p-r)^k + (8p-q)^k =

(-q)^k + (2p-r)^k + (3p-u)^k + (3p-v)^k + (5p+u)^k + (5p+v)^k + (6p+r)^k + (8p+q)^k

for k = 1,2,…7 where {p,q,r,u,v} are as defined above. (Update, 8/4/09): Alternatively, the system can be expressed in a more symmetric form,

[73, 58, 41] = [70, 65, 32, 15]

[85, 74, 61] = [87, 71, 56, 26]

[218, 167, 29] = [224, 107, 102, 65]

[351, 265, 221] = [336, 309, 169, 73]

[493, 335, 65] = [497, 297, 155, 16]

Some exhibit interesting side relations between its terms, such as the smallest one with x₁-x₂ = x₇, x₁-x₃ = x₆, though with appropriate substitutions, they are still not enough to reduce it to a parametric form. (End update.)

Note 4: If there is an initial soln for a certain n, is there always an infinite number for this n?

Note 5: A small variation of the method and which mostly yields the same results is to use different terms of F₁ which are to be multiplied by n and have their signs changed. Again let {a,b,c,d,e,f} = {p+qu, r+su, t+u, p-qu, r-su, t-u} but to solve the system,

-a+b+c = -d+e+f

a+b+nc = d+e+nf

a^k+b^k+c^k = d^k+e^k+f^k, k = 2

giving,

{t,s,q} = {-pq-rs, -n-q, (1-n)/2} and F₁ at k = 6 again reduces to the form,

(Poly11)u²+(Poly12) = 0

though now defined diffently,

Poly11:= (-3+n)(5-2n+n²)p + (3+n)(5+2n+n²)r

Poly12:= (-3+n)(5-2n+n²)p³ + (25-7n-5n²+3n³)p²r + (-25-7n+5n²+3n³)pr² + (3+n)(5+2n+n²)r³

As before, solve y² = – (Poly11)(Poly12) with trivial soln {p,r} = {-(n+3), n-3}. For ex, for comparison, again let n = 12,

y² = – 75(75p+173r)(225p³+881p²r+1159pr²+519r³)

with non-trivial soln {p,r} = {-39, 17} which has the same r and yields the same terms as in the first curve. To compare the two methods, for n = 12, 15, 21, 30, 33, 135, non-trivial points {p,r} are,

First method: {-104, 17}, {-183, 34}, {-53, 17}, {-24, 13}, {-137, 39}, {-229, 106}

Second method: {-39, 17}, {-65, 34}, {-24, 17}, {?, ?}, {-50, 39}, {-111, 106}

Notice that the value for r is the same. If the two methods give the same results, why is there apparently no small non-trivial soln for n = 30 using the latter one? Furthermore, Poly2 and Poly12 as cubic polynomials have different but related discriminants D_i being,

D₁ = (5-2n+n²)(5+2n+n²)

D₂ = (5-2n+n²)(1+n²)

respectively. Do the two methods then for the same n always give the same results?

Q: Any other forms for equal sums of sixth powers with six terms?

6.3 Seven terms

It is unknown if the unbalanced multi-grade eqn,

x₁^k+x₂^k+x₃^k = x₄^k+x₅^k+x₆^k+x₇^k, for k = 2,4,6

has non-trivial solns. Q: Anyone can provide one, preferably parametric?

(Update, 2/22/10): In response to a post in sci.math, Giovanni Resta made a search and found that there are no solns with terms < 845. James Waldby would later extend the search radius to < 1280 with still no soln. However, there are a few that are good only for k = 2,6, namely,

Note 3: Given a parametric soln to k = 1,2,6, it seems one can modify this to solve na+b+c = nd+e+f. For ex, using Choudhry’s soln to x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k, with {x_i, y_i} as,

{x₁, x₂, x₃} = { 2(α+β)m+(α-β+t), -2αm+(α+β+t), -2βm-(α+β-t)}

{y₁, y₂, y₃} = {-2(α+β)m+(α-β+t), 2αm+(α+β+t), 2βm-(α+β-t)}

discussed previously which is already true for k = 1,2, a linear combination of terms involving only the constants α,β should be found. Three possibilities are,

nx₁+x₂-x₃ = ny₁+y₂-y₃, x₁+nx₂-x₃ = y₁+ny₂-y₃, x₁-x₂+nx₃ = y₁-y₂+ny₃,

where n = (α-β)/(α+β), n = (α+2β)/α, and n = (α+2β)/β, respectively. If either α,β = 1, then the last two yield integer n. For α = 1, Choudhry found β = {7, 10, 16, 67} so using the second formula this gives n = {15, 21, 33, 135}, all n = 3m and consistent with the odd n found by this author.