025: Sixth Powers (8 terms)

Piezas

To find an infinite family of polynomial solns to k = 1,2,4,6, we will use 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

the big brother of F₁ (which was for six terms with k = 2,4 and k = 1,2,6). To make F₂ valid for k = 1,2, set g = -(ab+cd+ef) and f = -(1+b+d). Expanding for k = 4,6, we get,

(Poly11)h² + (Poly12) = 0 (eq.1)

(Poly21)h⁴ + (Poly22)h² + (Poly23) = 0 (eq.2)

respectively, where the Poly_i are expressions in the variables v_i = {a,b,c,d,e}. There are two ways to solve these two: first, to make Poly11 = Poly12 = 0, which will also cause eq.2 to factor into two quadratics, or second, to find v_i so eq.2 will factor into the same quadratic as eq.1. To find a common root between eq.1 and 2, it is a simple matter to eliminate the variable h using Mathematica’s Resultant[ ] function. (And since h is conveniently in even powers, it can be set h = √v to shorten the calculation.) The resultant is a factorable 6^th deg in e and, surprisingly, this has three non-trivial linear factors given by,

(a+ab-c+cd-be-de) (-a+ab+c+cd-be-de) (a+ab+c+cd-2e-be-de) = 0

plus an irreducible cubic factor. (One can get more solns if a linear root of the cubic can be found such as for special cases like d = -b when the above three become trivial but the cubic now has a single non-trivial linear factor.) In the general case though, solving for one appropriate variable of either of the three (b,d,e, respectively) and substituting into Poly11 = 0, one can solve for a second variable. These two can then set Poly11 = Poly12 = 0, and cause eq.2 to factor: two linear and one quadratic in h, with only the latter being non-trivial. Sparing the reader the rest of the algebra, given 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

where g = -(ab+cd+ef) and f = -(1+b+d), then for k = 1,2,4,6, if,

A. Linear factors:

First factor:

b = (1+d)(a+2e)/(a-e), c = (2+d)(a+e)/(1-d), where {a,d,e} satisfy

(Poly1)(-1+d)²h²-(Poly2)(a-e)² = 0

Poly1:= a²d-(9+20d+9d²)ae+de²; Poly2:= 9a²d-(1-20d+d²)ae+9de²

For arbitrary {a,d,e,h}, this already solves k = 1,2,4. For k = 6, the quadratic in h must be solved hence the problem is reduced to making its discriminant D a square, or D = (Poly1)(Poly2) = y². (Note that the Poly_i are curiously palindromic.) Since D is a quartic polynomial with a square leading and constant term, this is easily made a square. For example, using Fermat’s method, one soln is,

{a,e} = {9d(41+80d+41d²), -(5+4d)(4+5d)(20+59d+20d²)}

though presumably smaller polynomials may exist. Using this initial rational point on the curve D = y², subsequent ones can then be found.

Second factor:

a = (2+b)(c+e)/(1-b), d = (1+b)(c+2e)/(c-e), where {b,c,e} satisfy

(Poly3)(-1+b)²h²-(Poly4)(c-e)² = 0

Poly3:= bc²-(9+20b+9b²)ce+be²; Poly4:= 9bc²-(1-20b+b²)ce+9be²

so to solve (Poly3)(Poly4) = y². (After an exchange of variables this is essentially the same as in the first.)

Third factor:

b = -(a+2c)(a+c-2e)/(a²-c²+ae-ce), d = (2a+c)(a+c-2e)/(a²-c²+ae-ce), where {a,c,e} satisfy

(Poly5)h²-(Poly6)(a-c)²(a+c+e)² = 0

Poly5:= (a+c)²(2a+c)(a+2c)+(a+c)(a²-38ac+c²)e-(a²-38ac+c²)e²; Poly6:= (2a+c)(a+2c)+(a+c)e-e²

so to solve (Poly5)(Poly6) = y². This again has a square constant term so is easily made a square.

B. Alternative method

Still more solns can be found by solving Poly11 = Poly12 = 0 alternatively. Instead of eliminating h between eq.1 and 2, we eliminate the variable e between Poly11 and 12. This has seven non-trivial factors, with three essentially the same as the ones already discussed. The other four are the simpler,

(2+b+d)(1+b)(1+d)(b+d) = 0

The first is a special case of Poly11 = Poly12 = 0, giving a simple soln to 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

still with g = -(ab+cd+ef) and f = -(1+b+d), where,

{c,d,e,h} = {-ab(1-b)/w, -2-b, a(1-b)(3+2b)/w, ay/w},

with w = (2+b)(3+b) and b²+2b+21 = y².

This gives terms as quartic polynomials. (Alternatively, one can set d = 1 to get similar results.) The next three factors are variations on the same theme, namely a system that also satisfies x₁+x₂ = y₁+y₂ and x₃+x₄ = y₃+y₄ which imply an F₂ with d = -b; f = -1. Solving Poly11 = Poly12 = 0 gives a soln but is trivial. However, by using the previous method of eliminating h between eq.1 and eq.2, the resultant gives three factors (squared this time), the first two trivial but not the third. This, in fact, is the linear root of the irreducible cubic which reduces for the special case d = -b. Using this root, eq.2 factors into two quadratics, one identical to eq.1 so the problem is reduced to just solving this.

Thus, the complete soln to,

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

for k = 1,2,4,6 where x₁+x₂ = y₁+y₂ and x₃+x₄ = y₃+y₄ is given by 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

with {d, f} = {-b, -1} where {a,g,h} = {v₁(bc+e)e/w, v₂(bc+e)c/w, y/(2w)}, and w = v₁be+v₂c,

and {b,c,e} satisfying the quartic,

(2v₂c²-15bce+2v₁e²)² - v₁v₂c²e² = y²

where {v₁, v₂} = {b²-4, 1-4b²}. Since again this has a square leading and constant term, this is easily made a square with one small soln {c,e} = {8b, 1-4b²} giving,

{a,g,h} = {(b²-4)(1+4b²)/u, 8b(1+4b²)/u, (4-13b²+4b⁴)/u},

with u = b(b²+4) and yielding terms as 5^th degree polynomials in b. From this initial polynomial soln, one can then generate an infinite more. (In contrast to Crussol's approach which can only give infinite numerical solns.) If we are to consider the more general system,

nx₁+x₂+x₃+x₄ = ny₁+y₂+y₃+y₄

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

for k = 2,4,6, using our previous results for k = 1,2,4,6 it can be proven this has a soln for any n. We simply use the same method of changing the signs of appropriate terms since this does not affect the validity of the even powers. Thus, the system valid at k = 1,

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

where f = -1-b-d, is also good for,

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

if d is specialized as d = -2/(n+1). For example, an explicit soln to (Poly1)(Poly2) = y² for k = 1,2,4,6 (using the first factor) was given earlier as,

{a,e} = {9d(41+80d+41d²), -(5+4d)(4+5d)(20+59d+20d²)}

for any d. This can be used for k = 2,4,6 with (-x₁)+nx₂+(-x₃)+x₄ = (-y₁)+ny₂+(-y₃)+y₄, by letting d = -2/(n+1), hence the system can be solved for any n as claimed. (If n = -1, division by zero can be avoided by simply changing the sign of x₂,y₂ to change the sign of n.) Some concise examples for special n, derived slightly differently, are,

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

where a = 1, f = -1-bn-d, g = -(b+cd+ef), valid for k = 2,4,6 if,

1) {b,c,d,e,h} = {-2/(1+n), -2(-2+n)(3+n)/w, 2/(1+n), (1+n)(3+n)/w, (1+n)y/w}, where w = 2n(-5+n) and y² = 81-2n+n².

2) {b,c,d,e,h} = {-2/(1+n), -(3+n)(-1+3n)/w, -1, -2(3+n)/w, (1+n)y/w}, where w = 2n(1+3n) and y² = 21+38n+21n².

3) {b,c,d,e,h} = {-2/(1+n), -(3+n)(-2+3n)/w, -n/(1+n), -(3+n)/w, (1+n)y/w}, where w = 2n(-1+3n) and y² = -15+14n+33n².

which also are true for both,

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

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

Note 1: In contrast, as was seen, an infinite number of primitive solns to the system,

nx₁+x₂+x₃ = ny₁+y₂+y₃

x₁^k+x₂^k+x₃^k = y₁^k+y₂^k+y₃^k

for k = 2,6 and integer n,x_i,y_i can be proven only for some n which, other than the exceptional case n = 1, for the moment are all n = 3m. Whether there is an infinite number for some n ≠ 3m remains to be seen.

Note 2: If the terms are signed, all known solns to S₆ are valid for k = 1 as well. (In fact, also for S₈ and S₁₀.) Is there signed S₆ that is not for k=1? A simple computer search may be able to settle this.

Note 3: Lastly, as was mentioned, the system k = 1,2,4,6 with side conditions x₁+x₂ = y₁+y₂; x₃+x₄ = y₃+y₄ also has solns for just k = 1,2,6. The complete soln to this would be to ignore eq.1 and just solve eq.2 which, to recall, is a quartic with only even powers. To factor this into two quadratics entails the sufficient (but not necessary)¹ condition of making its discriminant D a square. As a polynomial in the variable a, D is a quartic with a square leading term so it is quite easy to do this. A simpler case is when b = -2 since the constant term of D also becomes a square. Set c = 1 without loss of generality to get,

D: = (3-4e+2e²)² + (-39+88e-64e²+16e³)a + (73-88e+28e²)a² + 3(-13+8e)a³ + 9a⁴

Three small non-trivial solns to D = y² are,

a₁ = 8(2e-e²)/(-17+16e), a₂ = -(21-22e+8e²)/(-19+8e), a₃ = (13-18e+8e²)/(-13+16e),

giving three families solving the specialized form of F₂ for just k = 1,2,6,

(a-2h)^k + (1+2h)^k + (e-h)^k + (g+h)^k = (a+2h)^k + (1-2h)^k + (e+h)^k + (g-h)^k

where g = -2+2a+e and either,

1) a = 8(2e-e²)/(-17+16e), h = y₁/(-17+16e), and y₁² = (17-23e+8e²)(17+13e+8e²)

2) a = -(21-22e+8e²)/(-19+8e), h = y₂/(-19+8e), and y₂² = -53+82e-67e²+32e³

3) a = (13-18e+8e²)/(-13+8e), h = y₃/(3(-13+8e)), and y₃² = 3(2873-7878e+8415e²-4096e³+768e⁴)

Footnote 1: While it is a reasonable assumption that P: = x⁴+bx²+c = 0 can be completely factored into two quadratics by finding b,c such that the expression D = b²-4c is a square, a second approach is to find b,c such that it factors as P = (x²+mx+n)(x²-mx+n) = 0 in which case it generally does not have a D that is a square.

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