023: Sixth Powers (4 or 6 terms)

Update (6/24/09): A. Choudhry

The fact that solving the systems S₁, S₂ is reducible to an elliptic curve implies that one soln can lead to another. Choudhry ("On Equal Sums of Sixth Powers", 1994) has given an explicit construction for this. Given the system,

x^k+y^k+z^k = u^k+ v^k+ w^k, k = 2,6 with,

x²+xu-u² = w²+wz-z²

y²+yv-v² = u²+ux-x²

z²+zw-w² = v²+vy-y²

Let {x,y,z,u,v,w} = {x₁, x₂, x₃, y₁, y₂, y₃} be a soln, then a new one is given by, {x,y,z,u,v,w} = {ap+x₁q, bp+x₂q, cp+x₃q, dp+y₁q, ep+y₂q, y₃q}, where {p,q} and {a,b,c,d,e} are,

{p,q} = {-a(2x₁-y₁)-b(2x₂+y₂)+d(x₁+2y₁)-e(x₂-2y₂), a²+b²-d²-e²-ad+be}

{a,b,c,d,e} = {(r₁²+2r₁r₂)s₂, (r₃²-2r₃r₄)s₁, s₁s₂, -(r₁²+r₂²)s₂, (r₃²+r₄²)s₁}

{s₁, s₂} = {r₂²+r₂r₁-r₁², r₃²+r₃r₄-r₄²}, and {r₁, r₂, r₃, r₄} = {x₁-y₃, x₃-y₁, x₂-y₃, x₃-y₂}.

Example: The initial soln {x₁, x₂, x₃, y₁, y₂, y₃} = {3, 22, -19, 23, 15, 10} gives {x,y,z,u,v,w} = {4513, -104, 4693, -2273, -5099, 3352} after removing the common factor 3*39200². (End note.)

The first, after a small adjustment, yields Delorme’s 5th-deg identity given below, while the second yields a different 5th deg. Delorme in his paper gave identities of deg n for all 4 ≤ n ≤ 11, except n = 6,10. Q: Can anyone give an identity with polynomials of deg n = 6?

Note: It can be shown that the system a^k+b^k+c^k = d^k+e^k+f^k, for k = 2,6 with,

a²+nad-d² = -(b²-nbe-e²)

b²+nbe-e² = -(c²-ncf-f²)

c²+ncf-f² = -(a²-nad-d²)

has non-trivial solns only for n = ±1. Proof: Using the general form and the method described above, and elimination of appropriate terms, the final resultant eqn has either trivial factors, or a quadratic factor solvable over √(-1), or if (n+1)(n-1) = 0. Since the first two can be disregarded then only the last applies, proving the assertion.

S. Brudno (1976), J. Delorme

The 4th-deg Brudno-Delorme identity also has the side condition 3a+b+c = d+e+3f, call this S₃. Let,

a^k+b^k+c^k = d^k+e^k+f^k, for k = 2,6 where,

{a,b,c} = {-n⁴-n³-5n²+8n+8, (n³+7n-2)(n+2), 3(3n²+2n+4)}

{d,e,f} = {(n²-n+3)(n+2)², -4n³-5n²-8n+8, -n⁴+n²+14n+4}

(modified to reduced the size of the coefficients), then these satisfy S₁, S₂, S₃. The smallest non-trivial example is n = -3 which, after removing common factors, yields {-23, 10, 15} = {3, 19, -22}, the example given earlier. Interestingly, it can be shown that this identity depends on the equation x²-6y² = z². This author found that some of the relations are enough to derive a version using: one eqn from S₂ (at k = 2), two from S₁, and S₃. Let,

(ad+b)^k + (c-2d)^k + (de+f)^k = (de-f)^k + (c+2d)^k + (ad-b)^k

and using this substitution also on the side conditions, one can derive the variables as,

{a,b,c,d,e,f} = {-4x+11y, y, 2(x-3y)(x-2y)+y², x-y, 2x-3y, 2x-5y}, where x²-6y² = 1.

One can easily solve for {x,y} in the integers as a Pell equation or, since the eqn is homogeneous, in the rationals. Q: Is it possible to solve S₂, S₃ without solving S₁?

J. Delorme

Delorme gave identities of deg n for all 4 ≤ n ≤ 11 , except n = 6,10, that solves S₁, S₂, one of which is the 5th deg,

{x₁, x₂, x₃} = {3n⁵+8n⁴+9n³-4n²-9n-2, -2n⁵-n⁴+12n³+13n²+4n-1, -n⁵-9n⁴-13n³-7n²-7n-3}

{y₁, y₂, y₃} = {2n⁵+9n⁴+4n³-9n²-8n-3, -3n⁵-7n⁴-7n³-13n²-9n-1, -n⁵+4n⁴+13n³+12n²-n-2}

Q: Any linear relations between the x_i and y_i?

Update (7/14/09): Since the x_i, y_i are polynomials in n, it turns out the problem is equivalent to finding six unknown rational constants p_i such that p₁x₁+p₂x₂+p₃x₃+p₄y₁+p₅y₂+p₆y₃ = m for some constant m, preferably m = 0, for all n. Expanding and collecting powers of n, set the coefficients, which are polynomials in the p_i, equal to zero. For this particular identity, we can't find p_i such that m = 0, but instead find m = 40 given by the linear relation,

5x₁-12x₂-7x₃-12y₁+5y₂+7y₃ = 40 (End note)

{p,t,s} = {2(3+3n+2n²)(3+2n+2n²), (3+3n+2n²)(3-2n+4n²), (1+n)(9+4n²)}

Theorem 2: (Delorme) The system S₂ implies S₁ if it is also the case that,

ad(a²-d²) + be(b²-e²) + cf(c²-f²) = 0

S. Brudno (1970); S. Brudno, I. Kaplansky (1974)

This is a simple example that solves both S₁ and S₂ ,

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

Note that for k = 2, one must solve a² = b²+c²+d². The complete soln for k = 2,6 is then,

{c,d} = {ax/(a²+9b²), 3bx/(a²+9b²)}

where {a,b,x} satisfies the elliptic curve,

(a²-b²)(a²+9b²) = x²

Labelled as x_i, y_i, the terms also satisfy x₁y₁(x₁²-y₁²) + x₂y₂(x₂²-y₂²) + x₃y₃(x₃²-y₃²) = 0. One small soln, among infinitely many, is {a,b} = {5/4, 1} which yields,

65^k + 52^k + 15^k = 36^k + 67^k + 37^k

Piezas

This is another simple example that solves both systems,

(ad+b)^k + (c-2d)^k + (de+f)^k = (de-f)^k + (c+2d)^k + (ad-b)^k, k = 2,6

where {a,b,c,d,e,f} = {-4x+11y, y, 2(x-3y)(x-2y)+y², x-y, 2x-3y, 2x-5y}, and {x,y} satisfies x²-6y² = 1.

It is quite interesting that this particular Pell equation appears in the context of 6th powers. However, since the eqn is homogeneous, one can just as well solve it in the rationals. This is also discussed several sections below.

A. Bremner, Piezas

It turns out the complete soln of S₁, S₂ can be given in terms of a quartic polynomial that is to be made a square, hence can be treated as an elliptic curve. R. Guy in the same book mentions that Bremner gave a method that can find all parametric solns. I don’t have access to this paper yet but after some experimentation found a procedure. It suffices to use S₂ and one of the eqns of S₁. Using the general form,

(p+q)^k + (r+s)^k + (t+u)^k = (t-u)^k + (r-s)^k + (p-q)^k

for k = 2,6 and one of the others, say c²+cf-f² = -(a²-ad-d²) with variables changed appropriately, the complete soln is then,

{q,u} = {-rs(p+2t)/w, rs(2p-t)/w}, where w = p²+t² and {p,r,s,t} satisfy,

((p²-11pt-t²)s² + w²)r² = (p²+pt-t²+s²)w²

(Perhaps not surprisingly, the form F₅:= x²+xy-y² appears again.) Thus, the problem is to find {p,t,s} such that the expression,

((p²-11pt-t²)s² + (p²+t²)²)(p²+pt-t²+s²) = y²

which is only a quartic in the variable s is a square, though some {p,t,s} are trivial. So, given an initial soln, this can be treated as an elliptic curve to generate more. For ex, let,

{p,t,s} = {1, n, n-1}

This is trivial with respect to the original sextic eqn S₂ but, using the same {p,t}, we can find more rational points s on the curve with the next one (using Fermat’s method) non-trivial of deg-18. There are also small non-trivial solns,

{p,t,s} = {-n-1, (n-1)(n-2), 5+n²}

{p,t,s} = {(-3+n)(2+n), 3+3n+2n², 3+6n+n²}

either one of which yields scaled versions of the 4th-deg Brudno-Delorme identity given below. Again this can be used to generate more rational points with the first one yielding a rational 18-deg. (Using the same {p,t}, there is another rational point s that is also only a quadratic, s = 3-2n+n² for the first and s = 9+4n+n² for the second, but these are trivial.) Other solns are,

{p,t,s} = {2(1+n)(-1-n+n²)(-1+n+n²), (1+n)(1-n+n²)(1-5n+n²), (-1+n)(1+2n+7n²+2n³+n⁴)}

(a²+ad-d²)^k + (b²-be-e²)^k + (b²+be-e²)^k + (c²-cf-f²)^k + (c²+cf-f²)^k + (a²-ad-d²)^k = 2(a^2k+b^2k+c^2k-d^2k-e^2k-f^2k), for k = 1,3

If the LHS is zero, then so is the RHS, which proves the theorem. (Masato Kuwata, Equal Sums of Sixth Powers and Quadratic Line Complexes, 2007. End proof.) (Note: This probably contributes to the reason why more than 90% of solns are multi-grade for k = 2,6 since being valid for one power means being valid for the other. Unfortunately, no similar expressions are yet known for 8th powers!)

Corollary 1: (Piezas) Solutions to S₁ imply the ff sums are squares,

4(a²+ad-d²) + 5b² = (b+2e)², 4(b²+be-e²) + 5c² = (c+2f)², 4(c²+cf-f²) + 5a² = (a+2d)²

4(a²-ad-d²) + 5c² = (c-2f)², 4(b²-be-e²) + 5a² = (a-2d)², 4(c²-cf-f²) + 5b² = (b-2e)²

In Unsolved Problems in Number Theory, R. Guy asked if the reverse was true: does S₂ imply S₁? This was answered in the negative by Delorme who gave one more condition, and Choudhry who gave a parametric example that solved S₂ but not S₁.

a) nx₁+x₂-x₃ = ny₁+y₂-y₃, for n = 12, 15, 21, etc.

If you know of another class, pls send it. The second condition of the first class can be expressed as a set of three simultaneous eqns given below. As was mentioned, small solns to (k.3.3) for k = 4 or 6 also turn out to be valid for k = 2. The smallest for k = 6 is,

23^k + 10^k + 15^k = 3^k + 19^k + 22^k

which, in fact, is for k = 2,6. This deceptively simple-looking eqn has a lot of structure. If expressed as,

{-23, 10, 15} = {3, 19, -22}

labelled {a,b,c,d,e,f} respectively, then,

a²+ad-d² = -(b²-be-e²)

b²+be-e² = -(c²-cf-f²)

c²+cf-f² = -(a²-ad-d²)

3a+b+c = d+e+3f

Not surprisingly, this is just the smallest instance of a parametric soln. Cubic and 4th powers have already been discussed and it was seen that a lot of identities involved equivalent forms of F₃:= x²+xy+y². For 5th and 6th powers, it seems now it is the form F₅:= x²+xy-y² that is implicit in some identities. For example, recall that,

(√p+√q)⁵ + (√p-√q)⁵ = (√r+√s)⁵ + (√r-√s)⁵

{p,q,r,s} = {5vw², -1+uw², 5v, -(u+10v)+w³}

where w = u²+10uv+5v² and it takes only a small change of variables {u,v} = {x-2y, x+2y} to transform this to the form x²+xy-y². Though F₃ and F₅ are similar-looking, they are quite different since the former has discriminant d = -3 and hence factors over the imaginary field √-3, while the latter has d = 5 and factors over the real field √5. It has been shown that solns to a²+ab+b² = c²+cd+d² also solve a^k+b^k+(a+b)^k = c^k+d^k+(c+d)^k for k = 2,4, and vice versa. For k = 2,6, one now needs a system of three equations,

Theorem 1: (Bremner, Kuwata) If there are {a,b,c,d,e,f} such that, call this system S₁,

a²+ad-d² + (b²-be-e²) = 0

b²+be-e² + (c²-cf-f²) = 0

c²+cf -f² + (a²-ad-d²) = 0

is true, then so is system S₂,

a^k+b^k+c^k = d^k+e^k+f^k, for k = 2,6

Proof: Simply add the expressions together to get the identically true statement,

2. x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k, for k = 1,2,6. Additional conditions include,