021: Fifth Powers (7 or 8 Terms)

a²+b²+c² = d²+e²+f² (eq.1)

abc = def (eq.2)

2) LHS: -p+2r = 0

for example, though one can choose any term on one side. Chernick's two identities have,

1) {p,q,r,u,v} = {c, b, d, 2a, 2c}

2) {p,q,r,u,v} = {d, b/2, d/2, a, c}

which, when substituted into the theorem, explains why one term is equal to zero.

J. Chernick, L. Lander

Theorem: "If a²+b²+c² = d²+e²+f², and abc = def, then,

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

For indefinite k, call this system C₁. Labeled as,

x₁^k + x₂^k + x₃^k + x₄^k = x₅^k + x₆^k + x₇^k + x₈^k, (eq.0)

note that this also satisfies x₁+x₂+x₃+x₄ = x₅+x₆+x₇+x₈ = 0. The history of C₁ is quite interesting. In 1913, Crussol considered a transposed form of C₁ to solve k = 1,2,4,6. In 1937, Chernick studied this system in the context of an ideal solution of degree 3 in standard form and gave the complete soln, though he knew by Theorem 6 it could be extended to fifth powers. In the late 1960s, Lander would revisit this system, with a small rearrangement of terms, focusing only on odd powers and solving the cases k = 1,5 and k = 1,3,5. Surprisingly, this system can be extended to solve seventh powers as well as was done by A. Choudhry in the 1990s to find the first numerical solns to k = 1,3,7 though the variables must fulfill a more complicated condition. Note that either side is a special case of Boutin’s Theorem,

(a+b+c)³ + (a-b-c)³ - (a+b-c)³ - (a-b+c)³ = 24abc

a theorem which generalizes the difference of two squares to other kth powers. Also, in the guise,

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

this appears in the Birck-Sinha Theorem which involves eighth powers. Whether C₁ has a soln for k = 1,8 or even k = 1,2,8 still remains to be seen, but perhaps there is. For k = 1,2,3,5, the problem then is to solve the two conditions,

a²+b²+c² = d²+e²+f² (eq.1)

abc = def (eq.2)

By adding the condition a±b±c = 0 (for any choice of signs), this implies one of the x_i of eq.0 will vanish and hence will involve only seven terms.

A. Partial Solutions

The section on Sums of n Squares gives a lot of identities for the first eqn such as the nice symmetrical one by Goldbach,

(p+r)² + (q+r)² + (p+q-r)² = (p-r)² + (q-r)² + (p+q+r)²

and it satisfies eq.2 as well if p²+pq+q² = r². Lander also gave the partial soln,

{a,b,c} = {q(pr+s), (p-r)(pr-s), pq(p+r)}

{d,e,f} = {q(pr-s), (p+r)(pr+s), pq(p-r)}

where s = q²-r². For the special case when one of the terms x_i vanishes, let one of the four (d±e±f) be equal to zero. It suffices to consider d+e±f = 0, hence,

q(pr-s) + (p+r)(pr+s) ± pq(p-r) = 0

Solving this quadratic in p, its discriminant must be made a square and involves an elliptic curve. Using the positive, then negative case yields,

5q⁴-8q²r²+4r⁴ = t₁²

-3q⁴+12q³r+4q²r²-8qr³+4r⁴ = t₂²

respectively, both of which have small solns, one being {q,r} = {4,1}, etc.

B. Complete Solution

Completely solving the two eqns,

1) RHS: 2p-v = 0

Since, after minor sign changes of p, the variables {q, r} are interchangeable (as well as the {u,v}), then there are two ways to set one term equal to zero, either:

(p+2q)^k + (p-2q)^k + (-p+2r)^k + (-p-2r)^k = (-2p+u)^k + (-2p-u)^k + (2p+v)^k + (2p-v)^k, for k = 1,2,3,5

with the second condition easily found by substituting c from eq.1 into eq.2, though trivial {a,b} should be avoided. Similarly for the second pair,

B. If (a+d)^k + (-a+d)^k + b^k = (c+d)^k + (-c+d)^k + d^k, for k = 2,4, then,

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

Let,

11a²+4b² = 9c² (eq.1)

2a²+b² = 2c²+d² (eq.2)

or alternatively,

11a²+4b² = 9c²

-4a²+b² = 9d²

which can be similarly treated as above. These are special cases of the general theorem given above. Recall that, for certain {p,q,r,u,v},

2a²+4b² = 9c²

14a²+b² = 9d²

This, in turn, can be completely solved in terms of binary quadratic forms as was already discussed in Fourth Powers (Part 3). Note that the one for k = 1,2,3,5 have sums of terms on either side equal to zero so, by Theorems 1 and 6 of the Prouhet-Tarry-Escott Problem, is essentially an ideal soln of degree 3 in standard form. The special case of eq.0,

x₁^k+x₂^k+x₃^k+x₄^k = x₅^k+x₆^k+x₇^k+x₈^k

when one term is equal to zero entails making a certain quartic polynomial a square (to be discussed later) and has has an infinite family of polynomial solns the smallest of which found by this author is of fifth degree, one by Chernick as ninth degree, though there are others also due to Chernick which need solving an elliptic curve but are notable due to their aesthetic form.

Chernick

One can derive a soln such that the same variables solve two sets of systems, for k = 2,4 and k = 1,2,3,5. There are two different pairs, special cases of the more general identity given above with the first being,

A. If (2a+c)^k + (-2a+c)^k + (2b)^k = (3c)^k + (-c)^k + (2d)^k, for k = 2,4, then,

(2a-2c)^k + (-2a-2c)^k + (4c)^k = (2b+c)^k + (-2b+c)^k + (-c-2d)^k + (-c+2d)^k, for k = 1,2,3,5.

This can be solved as,

2a²+4b² = 9c² (eq.1)

2a²+b² = 2c²+d² (eq.2)

Alternatively,

-3p²+q²+3r² = u²

-3p²+3q²+r² = v²