014: Sums of three Fourth Powers (Part 1)

Piezas

Theorem: If (p+u)^k + (p-u)^k + x^k = (p+v)^k + (p-v)^k + y^k, k = 2,4, then,

(2p+v)^k + (2p-v)^k + (-2p+u)^k + (-2p-u)^k = (p+x)^k + (p-x)^k + (-p+y)^k + (-p-y)^k, k = 1,2,3,5,

and where one of the terms of the latter eqn can be set to zero to get either of Chernick’s two solns.

Chernick, Escott

Theorem: If 1^k + 2^k + 3^k = u^k + v^k + (u+v)^k, k = 2,4, then,

(u-7)^k + (u-2v+1)^k + (3u+1)^k + (3u+2v+1)^k = (u+7)^k + (u-2v-1)^k + (3u-1)^k + (3u+2v-1)^k, k = 2,4,6.

Sinha

Theorem: If (a+3c)^k + (b+3c)^k + (a+b-2c)^k = (c+d)^k + (c+e)^k + (-2c+d+e)^k, k = 2,4, then,

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

excluding the trivial case c = 0. This was derived using the next theorem.

Birck, Sinha

Theorem: If a^k+b^k+c^k = d^k+e^k+f^k, k = 2,4 where a+b ≠ c; d+e ≠ f, then,

(a+b+c)^k + (-a+b+c)^k + (a-b+c)^k + (a+b-c)^k + (2d)^k + (2e)^k + (2f)^k =

(d+e+f)^k + (-d+e+f)^k + (d-e+f)^k + (d+e-f)^k + (2a)^k + (2b)^k + (2c)^k

for k = 1,2,4,6,8.

Piezas

Theorem: If a^k+b^k+c^k = d^k+e^k+f^k for k = 2,4. Define {x,y} = {a²+b²+c², a⁴+b⁴+c⁴}. Then,

a⁶+b⁶+c⁶-d⁶-e⁶-f⁶ = 3(a²-d²)(b²-d²)(c²-d²) = 3(a²b²c²-d²e²f²)

3(a⁸+b⁸+c⁸-d⁸-e⁸-f⁸) = 4(a⁶+b⁶+c⁶-d⁶-e⁶-f⁶)(x)

6(a¹⁰+b¹⁰+c¹⁰-d¹⁰-e¹⁰-f¹⁰) = 5(a⁶+b⁶+c⁶-d⁶-e⁶-f⁶)(x²+y)

If we define F_k:= a^k+b^k+c^k-d^k-e^k-f^k, then for even k > 4, F_k is of the form F₆Poly1 and also a highly composite number. More significantly, the identities can be combined and for a,b,c,d,e,f with appropriate properties, these can be woven together into a very neat form. For example, define n as,

a⁴+b⁴+c⁴ = n(a²+b²+c²)²

and Q₁ implies the general identity,

32(a⁶+b⁶+c⁶-d⁶-e⁶-f⁶)(a¹⁰+b¹⁰+c¹⁰-d¹⁰-e¹⁰-f¹⁰) = 15(n+1)(a⁸+b⁸+c⁸-d⁸-e⁸-f⁸)²

For the special case when a+b = ±c, since,

a⁴+b⁴+(a+b)⁴ = (1/2)(a²+b²+(a+b)²)²

then n = 1/2, and the above becomes,

64(a⁶+b⁶+c⁶-d⁶-e⁶-f⁶)(a¹⁰+b¹⁰+c¹⁰-d¹⁰-e¹⁰-f¹⁰) = 45(a⁸+b⁸+c⁸-d⁸-e⁸-f⁸)²

which is a concise version of Ramanujan’s 6-10-8 Identity discussed in the next section. (In the general case, this turns out to be just the first in a family of identities.) (Update, 7/31/09): This can be extended to the "solution chain" a₁^k+a₂^k+a₃^k = a₄^k+a₅^k+a₆^k = a₇^k+a₈^k+a₉^k = a₁₀^k+a₁₁^k+a₁₂^k = ..., k = 2,4, for a chain divisible into pairs. For the special case when {a₁+a₂, a₄+a₅, a₇+a₈, a₁₀+a₁₁} = {a₃, a₆, a₉, a₁₂}, then,

64(a₁⁶+a₂⁶+a₃⁶+a₄⁶+a₅⁶+a₆⁶-a₇⁶-a₈⁶-a₉⁶-a₁₀⁶-a₁₁⁶-a₁₂⁶)(a₁¹⁰+a₂¹⁰+a₃¹⁰+a₄¹⁰+a₅¹⁰+a₆¹⁰-a₇¹⁰-a₈¹⁰-a₉¹⁰-a₁₀¹⁰-a₁₁¹⁰-a₁₂¹⁰)

= 45(a₁⁸+a₂⁸+a₃⁸+a₄⁸+a₅⁸+a₆⁸-a₇⁸-a₈⁸-a₉⁸-a₁₀⁸-a₁₁⁸-a₁₂⁸)²,

One can test this by the soln chain [28, 175, 203] = [77, 140, 217] = [107, 113, 220] = [5, 188, 193] given here, found by Gloden in the 1940's. (End update.) It seems then the system Q₁ is quite important and worth a second look. To solve,

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

for k = 2,4, one trick is to reduce the degree of the two eqns by using the substitution {a,b,c,d,e,f} = {p+q, r+s, t+u, t-u, r-s, p-q} to get,

pq+rs+tu = 0

p³q+pq³+r³s+rs³+t³u+tu³ = 0

By eliminating u between them, one ends up with a quartic in t,

(pq+rs)t⁴-(p³q+pq³+r³s+rs³)t²+(pq+rs)³ = 0

or a cubic in either p,q,r,s. However, it is possible to use another substitution that can reduce the degree even further.

Theorem: "The system a^k+b^k+c^k = d^k+e^k+f^k for k = 2,4, call it Q₁, can be reduced to solving a quadratic. Its discriminant D is a quartic polynomial in one variable and the problem of making D a square can be treated as an elliptic curve."

(In fact, as was already mentioned, similar results can be given for the three systems k = 1,4, or k = 1,5, or k = 1,2,6. Another substitution will be used for these which is slightly more efficient since the first system ends up with a discriminant D that is only a quadratic polynomial.)

Proof: We simply use a different substitution, call this form F₁,

{a,b,c,d,e,f} = {p+qu, r+su, t+u, t-u, r-su, p-qu} to get,

pq+rs+t = 0

p³q+r³s+t³+(pq³+rs³+t)u² = 0

where the degree of the variable u has been reduced to merely a quadratic. Eliminating t between them yields,

(Poly1)u²-(Poly2) = 0

where,

Poly1:= p(q-q³)+r(s-s³)

Poly2:= p³(q-q³)-3pqrs(pq+rs)+r³(s-s³)

and the problem is reduced to making its discriminant a square,

y² = (Poly1)(Poly2)

which is just a quartic in p (or r), thus proving the theorem. This quartic in fact has a square leading and constant term so is easily made a square and from that initial point, other rational ones may be computed. To prove that this is the complete characterization of non-trivial solns, solve for p,q… in terms of the original variables to get,

{p,q,r,s,t,u} = {(a+f)/2, (a-f)/(c-d), (b+e)/2, (b-e)/(c-d), (c+d)/2, (c-d)/2}

and even if there is division by a variable, when c = d the system a^k+b^k = e^k+f^k for k = 2,4 has only trivial solns hence this exceptional case is of no interest and does not affect the generality of the theorem. While it has been proven that Q₁ can be reduced to solving a quadratic, a third constraint may be useful to explore various solns. This constraint is an appropriate linear or quadratic relation between the terms such that the final eqn remains a quadratic and involve a simpler elliptic curve. This author found nine relations, six linear and three quadratic,

10.1 a+b = nc; d+e = nf

10.2 a+b ≠ c; d+e ≠ f

10.3 a+b±c = n(d+e±f)

10.4 na+b+c = d+e+nf

10.5 na+b = e+nf

10.6 a+d = n(c+f)

10.7 (a²-f²)c² = -(b²-e²)d²

10.8 a²+nab+b² = d²+nde+e²

10.9 a²+nad+d² = b²+nbe+e²

for any n. All of these have a complete soln in terms of an elliptic curve though may simplify into quadratic forms for particular values of n or be trivial when n = 0. The linear relations will be discussed in the next section.

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