026: Seventh Powers (8, 9, 10 terms)

II. Constraint: x₁+x₂ = x₃+x₄ = y₁+y₂ (where x₁+x₂+x₃+x₄+x₅ = y₁+y₂+y₃+y₄+y₅)

2. (6x+3y)^k + (4x+9y)^k + (2x-12y)^k = (-x+3y+3z)^k + (-x+3y-3z)^k + (-6x-6y)^k

where x²+10y² = z², D = -10. (This one, after minor changes, gives the example at the start of this section.)

where D is the discriminant of the binary quadratic form. Furthermore, treated as rational points on a certain elliptic curve, it can be showed that an infinite sequence can be derived from these thus providing polynomial solns of increasing degree. (See discussion on Fourth Powers.) The example below is based on Sinha’s soln [1],

(x-y+z)^k + (-3x+y+z)^k + (x-y-z)^k + (-3x+y-z)^k + (14x-y)^k =

(4x-3y)^k + (-y)^k + (11x-2y)^k + (x+2y)^k + (-6x+3y)^k

for k = 1,3,5,7, if 126x²-5y² = z².

Note: To compare, for eighth powers, other than the Letac-Sinha identity, there are only two solns known found by computer search with all terms x_i < 520. One obeys the same constraints as Sinha’s identity, x₁-x₂ = x₃-x₄ = y₁-y₂, (and an additional x₄-x₂ = y₄-y₂). This is,

[366, 103, 452, 189, -515] = [508, 245, -18, 331, -471], for k = 1,2,4,6,8

but whether this has a parametric soln is unknown.

1. (-5x+2y+z)^k + (-5x+2y-z)^k + (6x-4y)^k = (9x-y)^k + (-x+3y)^k + (16x-2y)^k

where 126x²-5y² = z², D = 9(70). (Sinha’s)

It can be shown that given the system k = 1,3,5,7,

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

if it has either of the constraints,

1) x₁-x₂ = x₃-x₄ = y₁-y₂

2) x₁+x₂ = x₃+x₄ = y₁+y₂

(and equivalent forms after transposition such as the one above), then an infinite sequence of polynomial solns can be found by finding non-trivial rational points on a quartic polynomial that is to be made a square.

I. Constraint: x₁-x₂ = x₃-x₄ = y₁-y₂ (where x₁+x₂+x₃+x₄+x₅ = y₁+y₂+y₃+y₄+y₅)

Here is a beautiful theorem by T. Sinha. (See update in next page, bottom section.) If,

(a+3c)^k + (b+3c)^k + (a+b-2c)^k = (c+d)^k + (c+e)^k + (-2c+d+e)^k, for k = 2,4,

then,

a^k + (a+2c)^k + b^k + (b+2c)^k + (-c+d+e)^k = (a+b-c)^k + (a+b+c)^k + d^k + e^k + (3c)^k, for k = 1,3,5,7,

excluding the trivial case c = 0. Note that this also satisfies,

x₁-x₂ = x₃-x₄ = y₁-y₂ = -2c,

2(x₁+x₃) = y₁+y₂

x₅ = y₃+y₄-(1/3)y₅

a set of conditions which is enough to give the complete soln. An example is,

{a,b,c,d,e} = {43, 9, 1, 19, 37}

yielding,

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

which is the smallest solution mentioned previously. Sinha’s problem then is to solve,

p₁^k+p₂^k+p₃^k = q₁^k+q₂^k+q₃^k

for k = 2,4 where p₁+p₂-p₃ = 2(q₁+q₂-q₃) ≠ 0. He gave a single polynomial soln to this in terms of binary quadratic forms but this author showed there are two such solns, though there could be more.

(Update, 1/25/10): Piezas

Previously, it was seen how a k = 1,3,5 can lead to a k = 2,4,6. This can be brought higher. Given a (k.4.4) for k = 2,4,6, if its terms {x_i , y_i} are such that x₁-x₂ = y₁-y₂ = y₂-y₃ , then it can lead to a (k.5.5) for k = 1,3,5,7. This is exemplified by the infinite family,

(3a+2b+c)^k + (3a+2b-c)^k + (-2a-b+d)^k + (-2a-b-d)^k = (3a+b+2c)^k + (3a+b)^k + (3a+b-2c)^k + (-7a-b)^k, for k = 1,2,4,6

which leads to the higher system,

[3a+b+3c, 3a+2b-2c, -7a-b+c, -2a-b-c-d, -2a-b-c+d]^k =

[3a+b-3c, 3a+2b+2c, -7a-b-c, -2a-b+c-d, -2a-b+c+d]^k, for k = 1,2,3,5,7

where, for both, 10a²-b² = c², and 55a²-6b² = d², with trivial ratios a/b = {1, 1/3, 5/13}. It is tempting to speculate that there may be higher-order version of this for (k.5.5), for k = 2,4,6,8. (End update.)

From Choudhry's result, we know H₄ can have non-trivial rational roots, one of which is h = 98 when {a,c,e,g} = {3166, 2273, 3378, 1839}. To find a polynomial identity, as was pointed out in the previous section, there are two ways to factor quartics with only even exponents, one of which is to make its discriminant a square. One soln is a+c+e+g = 0. Let {a,c,e,g} = {p+q+r, p-q-r, -p+q-r, -p-q+r} for symmetry, and H₄ has the non-trivial quadratic factor,

3h²(q²+r²) + (5p²+3q²+3r²)(q+r)² = 0

Unfortunately, this can only be solved by complex values. If we let h = hi where i is the imaginary unit, then we are to find,

3(5p²+3q²+3r²)(q²+r²) = y²

As it is only a quadratic in p with a square constant term, this is easily done. In summary, for the moment, only particular values (extrapolated from Choudhry's cubic) are known such that H₄ has a non-trivial rational h. If an appropriate general relation can be found between {a,c,e,g} such that H₄ has a quadratic factor with real roots, then it may be possible to find an infinite family of polynomials for k = 1,3,7, just as there is one for k = 1,5.

Note: For k = 1,3,5,7 with ten terms, a similar situation arose, a parametrization dependent on a multi-variable quartic polynomial with only even exponents. However, this author was able to find two relations between the variables such that the quartic factored. See discussion in the subsequent sections.

7.2 Nine terms

If x₁ = 0, there are only two known primitive solns to,

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k, for k = 1,3,5,7

These lead to ideal solns of deg 8. Interestingly, terms can be transposed such that x₁+x₂+x₃+x₄+x₅ = y₁+y₂+y₃+y₄+y₅ = 0. (There is a situation similar for k = 1,3,5 wherein the only known solns with x₁ = 0 can be transposed in the same manner. It is interesting to consider what the first k = 1,3,5,7,9 with x₁ = 0 would look like.) Letac in 1942 found the two,

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

[0, 63, 119, 161, 169] = [8, 50, 132, 148, 174]

(How did he find it prior to computer searching?) (Update, 8/15/09): Christian Bau did a search for terms < 200 and James Waldby extended it to < 450 with no new primitive solns being found.) It seems unknown if there is a parametric soln. But there are two points to be noted. First, as pointed out by Bailey et al, they can be transposed so an equal number of terms on either side sum to zero. For the first, it is,

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

The second one has more "structure" as it can be re-arranged in four distinct ways,

[0, 63, 119, -8, -174] = [-161, -169, 148, 50, 132]

[0, 63, 119, -50, -132] = [-161, -169, 148, 8, 174]

since 8+174 = 50+132, and,

[63, -50, -174, 161, 0] = [-119, 132, 148, -169, 8]

[-119, 132, 148, -161, 0] = [63, -50, -174, 169, -8]

since 169-8 = 161-0. As the left hand side involves only four terms, this (and only this side) also obeys,

5(x₁²+x₂²+x₃²+x₄²)(x₁³+x₂³+x₃³+x₄³) = 6(x₁⁵+x₂⁵+x₃⁵+x₄⁵)

a general identity first mentioned in k = 1,3,5. Second point, for the first soln, re-arranged as,

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

Shuwen found this is good for one even exponent, k = 2. Later we shall see that both these phenomena appear in the only (and presumably smallest) soln for k = 1,3,5,7,9.

Piezas

Conjecture: "Define F_k:= x₁^k+x₂^k+x₃^k+x₄^k-(y₁^k+y₂^k+y₃^k+y₄^k+y₅^k) where x₁+x₂+x₃+x₄ = y₁+y₂+y₃+y₄+y₅ = 0. If F_k = 0 for k = 1,3,5,7, then F₂² = -2F₄."

This works for the two known solns. I don’t know whether this is true in general for systems of this form or is just a peculiarity of Letac’s method.

7.3 Ten terms

The next step is S₇ and S₈,

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

for k = 1,3,5,7, or k = 2,4,6,8. It can be proven by computer search that the smallest solution to the first system, at least in distinct integers, is given by,

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

This turns to have already been known (pre-computer era) by Sinha as belonging to an infinite family. However, there are just a few methods known to parametrically solve S₇ (at least previously) and just one for S₈, which is deplorable. With the advent of computer algebra systems, the Internet, and hopefully more people working on the field of equal sums of like powers (ESLP), perhaps this can be rectified eventually. Ironically, compared to S₆, it seems solns to S₇ are rather plentiful since while only one binary quadratic form identity (Chernick’s) is known for the former, three are known for the latter. (Then again, it could be merely a lack of knowledge about S₆.)

1st: The soln given by Sinha,

(x-y+z)^k + (-3x+y+z)^k + (x-y-z)^k + (-3x+y-z)^k + (14x-y)^k =

(4x-3y)^k + (-y)^k + (11x-2y)^k + (x+2y)^k + (-6x+3y)^k

for k = 1,3,5,7, if 126x²-5y² = z², D = 9(70).

2nd: One soln found by this author is the beautifully simple Sagan's Identity,

1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = (-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k

for k = 1,3,5,7 if x²-10y² = 9.

Note that this also satisfies x₁+x₂ = x₃+x₄ = -(x₅+x₆) as well as some other obvious ones. While the complete soln uses rational x,y, if the condition is treated as a Pell eqn, then there are integral ones giving an infinite number of solns where two terms {1, 5} are constants.

Update, 11/7/09: Last year, I dedicated an algebraic identity I found to the astronomer and science writer Carl Sagan (1934-1996). I read his "Broca's Brain" and "The Dragons of Eden" in my late teens, and read and eventually saw "Contact". Since I already dedicated one article I wrote on Degen's Eight-Square Identity to the novelist Katherine Neville, author of the amazing book "The Eight", I thought it was only fitting I name one of the identities I found after Sagan. After all, it has billions and billions of solutions. (You need to be a Sagan fan to understand the previous remark.) Today is the first Carl Sagan Day, and I decided to revive the article I wrote about Sagan's Identity that perished when Geocities was closed by Yahoo last Oct. 26. (End update.)

3rd: Also by this author,

(3x+2y)^k + (-3x+7y)^k + (3x-2)^k + (3+2x)^k + (-5x-3y)^k =

(-3x+2y)^k + (3x+7y)^k + (-3x-2)^k + (3-2x)^k + (5x-3y)^k

for k = 1,2,3,5,7 if x²+6y² = 1 which has the distinction of being valid for k = 2 as well.

All known parametrizations for k = 1,3,5,7 in terms of binary quadratic forms involve the unsigned square-free discriminants D = 6, 10, 70. In short, the set D only employs the small prime factors p = 2,3,5,7. Whether this is coincidence I do not know.

H₄: = 3(v₁⁴-v₃⁴)h⁴ + 5(v₁²v₄-v₂v₃²)h²v₁² + (a²c²-e²g²-v₂²+v₄²)v₁⁴ = 0

where {v₁, v₂, v₃, v₄} = {a²-c², a²+c², e²-g², e²+g²}.

a form which gives x₁+x₂+x₃+x₄ = y₁+y₂+y₃+y₄ = 0. Note that the x_i (or y_i) also satisfies,

5(x₁²+x₂²+x₃²+x₄²)(x₁³+x₂³+x₃³+x₄³) = 6(x₁⁵+x₂⁵+x₃⁵+x₄⁵)

Theorem 1: Eq. 1 is true for k = 1,3,7 if it satisfies the two conditions,

abc = def

2(a⁴+b⁴+c⁴) - 5(a²+b²+c²)² = 2(d⁴+e⁴+f⁴) - 5(d²+e²+f²)²

This is the same form used by Chernick for ideal solns of k = 1,2,3 and Lander for fifth powers k = 1,3,5, though as one can see the condition is a little more complicated. For the special case d+e = f, the system simplifies to satisfying the two eqns,

abc = de(d+e)

2(a⁴+b⁴+c⁴) - 5(a²+b²+c²)² = 16(d²+de+e²)²

though whether this has non-trivial solns in the rationals is not known. If there is, then one of the terms x_i or y_i vanishes so would be the first counter-example to Euler’s conjecture for seventh powers. However, if this constraint is not given, after a clever substitution which reduced the problem to finding a rational root of a cubic, Choudhry found several solns to Theorem 1, one is {a,b,c,d,e,f} = {324, 5439, 893, 5217, 1539, 196} which gives,

[3328, 2111, -3004, -2435] = [3476, 1741, -3280, -1937]

which is for k = 1,3,7 as well as satisfying x₁+x₂+x₃+x₄ = y₁+y₂+y₃+y₄ = 0, and 5(z₁²+z₂²+z₃²+z₄²)(z₁³+z₂³+z₃³+z₄³) = 6(z₁⁵+z₂⁵+z₃⁵+z₄⁵) where z_i = x_i, (or z_i = y_i).

Note 1: Kuosa's 1999 soln for k = 1,3,7 does not belong to this family as there is no transposition of four appropriate x_i such that their sum is equal to zero.

Note 2: Theorem 1 can also be satisfied by a^k+b^k+c^k = d^k+e^k+f^k, for k = 2,4 where abc = def. But, using resultants, it is easily proven that this restricted case has only trivial solns.

Piezas

Express eq.1 in the form F₂,

(a+bh)^k + (c+dh)^k + (e+fh)^k + (g+h)^k = (a-bh)^k + (c-dh)^k + (e-fh)^k + (g-h)^k

To satisfy the side condition x₁+x₂ = y₁+y₂; x₃+x₄ = y₃+y₄ which is essentially the same as Choudhry’s, let {b,f} = {-d, -1}. This makes it true for k = 1. Expanding for k = 3, we find the eqn,

d = -(e²-g²)/(a²-c²)

Substituting this into k = 7 we get a quartic in h with only even exponents, call this H₄, with coefficients in {a,c,e,g} as,

{x₁, x₂, x₃, x₄} = {a+b+c, a-b-c, -a-b+c, -a+b-c}

{y₁, y₂, y₃, y₄} = {d+e+f, d-e-f, -d-e+f, -d+e-f}

Given, x₁^k + x₂^k + x₃^k + x₄^k = y₁^k + y₂^k + y₃^k + y₄^k, (eq.1)

Choudhry found a multi-variable cubic polynomial such that when it has a rational root, it yields a k = 1,3,7. While no polynomial soln is yet known, it can be shown that if x_i and y_i are complex numbers with integer components, then there are families of polynomial solns (to be given later).

A.J. Choudhry

c) x₁⁷ + x₂⁷ + x₃⁷ + x₄⁷ + x₅⁷ + x₆⁷ + x₇⁷ = 0

Still unknown. (Though since five 5th powers can sum to both zero and 1, it may be a general property of odd n nth powers for n > 3.)