10. Form: x⁴+y⁴+z⁴ = t⁴

 

This equation is significant since Euler made the rather reasonable conjecture that it would take at least k kth powers to sum up to a kth power which turned out to be erroneous.  The first counter-example was found by Lander, Parkin as,

 

27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵

 

via computer search but there is no systematic method to find other primitive solns for fifth powers.  For fourth powers, a counter-example was later found as well as a method to generate further solns, albeit they can quickly become enormous.

 

Demjanenko, Elkies

 

Let {a,b} = {2m²+n²,  2m²-n²} for relatively prime integers m,n with n odd.  If there are rational {x,y,z} such that they satisfy,

 

ay² = (8mn-3a)x²-2bx-2mn                   (eq.1)

±az² = 4bx²+8mnx-b                            (eq.2)

 

then {x,y,z} gives a rational soln to,

 

(x+y)⁴ + (x-y)⁴ + z⁴ = 1

 

The technique starts by using appropriate {m,n} where n must be odd, and finding initial rational solns {x₁,y₁} to (eq.1).  This can yield a parametrization for the variable x as a quadratic polynomial and, by substituting this into (eq.2), now results in a quartic polynomial that is to be made a square.  For ex, for {m,n} = {8,-5}, (eq.1) is,

 

153y² = -779x²-206x+80

 

which has initial soln {x₁,y₁} = {3/14, 1/42}.  This yields a parametrization for x as,

 

x = (51p²-34p-5221)/(238p²+10906)

 

which, when substituted into (eq.2), becomes the problem of finding a value p such that the quartic polynomial is a square.  Clever transformations may then reduce the size of the coefficients.  Note also that the two conditional equations are quadratics in x and, by solving this, its discriminant D_i is in terms of {m,n,y² (or ±z²)}.  This author found that for eq.1 which has,

 

D₁ = 4b²-4(3a-8mn)(2mn+ay²)

 

where {a,b} = {2m²+n²,  2m²-n²}, it is possible to express {m,n,y²} as polynomials such that this is a square by using either of two solns,

 

{m,n} = {56v⁴+32v²-25,  96v⁴+120v²}

{m,n} = {48v⁴+60v²,  56v⁴+32v²-25}

 

where, for both, {v,y²} are related by v²+3y² = 1 which can be completely solved.  By Elkies’ result, only odd n are appropriate so only the second soln can be used.  These values still have to be substituted into eq.2 which unfortunately results in a polynomial of high degree that is to be made a square.

 

Elkies

 

(85v²+484v-313)⁴ + (68v²-586v+10)⁴ + (2u)⁴ = (357v²-204v+363)⁴

 

if u² = 22030+28849v-56158v²+36941v³-31790v⁴,

 

a soln of which is v = -31/467.  This gave the first rational soln to x⁴+y⁴+z⁴ = t⁴ (eq.0) and is just the case {m,n} = {8,-5} of the previous identity.  This elliptic curve in fact has an infinite number of rational points, thus providing an infinite number of solns to eq.0.

 

 

11. Form: x⁴+y⁴+z⁴ = nt^k

 

E. Fauquembergue

 

(ab)⁴ + (ac)⁴ + (bc)⁴ = (a⁴+a²b²+b⁴)²,  if a²+b² = c²

 

This beautifully simple soln has a counterpart a⁴+b⁴+c⁴+d⁴ = t² which also depends on Pythagorean triples and again found by Fauquembergue.  Any soln for k > 2?

 

F. Proth (and others)

 

a^k + b^k + (a+b)^k = 2(a²+ab+b²)^k/2,  for k=2,4

 

As was already pointed out, this is quite a ubiquitous algebraic form.  Its first even kth powers can be expressed as a sum of three squares,

 

(ab)² + (ab+b²)² + (a²+ab)² = (a²+ab+b²)²,

 

E. Escott

 

(ab)⁴ + (ab+b²)⁴ + ((a²+ab)(a²+ab+2b²))² = (a²+ab+b²)⁴,

 

In fact, these imply that the form (a²+ab+b²)^2k is expressible as the sum of three squares using the method discussed below.

 

Ramanujan

 

Let a+b+c = 0, then,

 

a⁴ + b⁴ + c⁴ = 2(ab+ac+bc)²

 

a⁴(b-c)⁴ + b⁴(a-c)⁴ + c⁴(a-b)⁴ = 2(ab+ac+bc)⁴

 

(a²b+b²c+c²a)⁴ + (ab²+bc²+ca²)⁴ + (3abc)⁴ = 2(ab+ac+bc)⁶

 

In his Notebooks, Ramanujan gave these as well as one for k=8 and wrote “…and so on”.  The rest can be found by noting that the first is equivalent to Proth’s so what one has to do is to find expressions a,b such that a²+ab+b² = (p²+pq+q²)^k which can be done by factoring over a complex cube root of unity ω,

 

(a-bω) (a-bω²) = (p-qω)^k (p-qω²)^k

 

and by using the reliable method of equating factors, one can then easily solve for {a,b}.  It should be pointed out Ramanujan may have used another method to derive his more elegant expressions which are conditionally dependent on a+b+c = 0.  (Incidentally, this form has a generalization for third and fifth powers.)

 

S. Realis

 

x⁴+y⁴+z⁴= 3t²

 

{x,y,z} = {5a⁴+4a³+9a²+10a+5,  5a⁴+10a³+9a²+4a+5,  5a⁴+16a³+27a²+16a+5}

 

Any other solns to x⁴+y⁴+z⁴= nt^k for n>2?

 

E. Fauquembergue

 

(2p²-2q²)⁴ + (p²-4q²)⁴ + (3pq)⁴ = (p²+2q²)⁴ + (4p⁴-11p²q²+16q⁴)²

 

 

12. Form: a^k+b^k+c^k = d^k+e^k+f^k  (k = 2,4)

 

To recall, a special case of Bastien’s Theorem would state that the above has no non-trivial soln for k = 1,2,3,4.  However, this does have solns when k = 2,4 or 1,2,4 or even 2,3,4 and, of course, when only for single exponents.  When valid for both k = 2,4, this quadratic-quartic system will be called throughout this work as Q₁, and this form is significant because it may appear as a side condition for higher powers such as for the fifth, sixth, seventh, and eighth.  First, we have these general theorems,

 

T. Sinha

 

Theorem:  If a^k+b^k+c^k = d^k+e^k+f^k,  k = 2,4, then (a²-f²)(b²-f²) = (c²-d²)(c²-e²).

 

The eqn (a²-f²)(b²-f²) = (c²-d²)(c²-e²) is quite easy to satisfy and can be reduced to solving an eqn of form x²+ny² = z².  However, it is not enough to imply Q₁.  More generally, any two of the equations in the system implies the third, but it is not the case that one implies the other two.

 

Tarry, Escott

 

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

 

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

 

k = 1,2,3,4,5 for any t.

 

This is a special case of a theorem of the Prouhet-Tarry-Escott Problem which I’m assuming is due to Tarry or Escott since they established related theorems for this field.  Note that appropriate pairs have the common sum 2t and is known as a symmetric solution.  Any multi-grade k = 1-5 of this form is then reducible to Q₁.  By a theorem due to Bastien, the system,

 

x₁^k+x₂^k+…+ x_m^k = y₁^k+y₂^k+…+ y_m^k

 

for k = 1,2,…n has non-trivial solns only if m>n.  The minimal soln is then m = n+1 and for the case k = 1,2,…5 this implies an eqn with 12 terms.  However, it is possible some terms on one side are equal to zero.  For example, given the special case of Q₁,

 

a^k+b^k+c^k = 2d^k+e^k,  k = 2,4, then

 

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

 

which used t = -d to yield this 10-term soln.

 

Chernick

 

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

 

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

 

which is one of two identities.  Note that if the terms of Q₁ are expressed as x_i and y_i, this also satisfies x₁+x₂ = y₁+y₂.  More generally,

Piezas

 

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

 

(2p+q+r)^k + (2p-q-r)^k + (-2p-q+r)^k + (-2p+q-r)^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(a⁸+b⁸+c⁸-d⁸-e⁸-f⁸) = 4(a⁶+b⁶+c⁶-d⁶-e⁶-f⁶)(a²+b²+c²)

6(a¹⁰+b¹⁰+c¹⁰-d¹⁰-e¹⁰-f¹⁰) = 5(a⁶+b⁶+c⁶-d⁶-e⁶-f⁶)((a²+b²+c²)²+a⁴+b⁴+c⁴)

 

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 ten relations, five linear and five quadratic,

 

1. a+b = nc; d+e = nf
2. a+b+c = n(d+e+f)
3. na+b+c = d+e+nf
4. na+b = e+nf
5. a+d = n(c+f)
6. (a²-f²)c² = -(b²-e²)d²
7. a²+nae-e² = -(b²+nbf-f²)
8. a²+nae+e² = b²+nbf+f²
9. a²+nab-b² = -(e²+nef-f²)
10. a²+nab+b² = e²+nef+f²

for any n.  All of these have a complete soln in terms of an elliptic curve though it may be trivial when n = 0 or may simplify for particular values such as n = 1.  Most will be discussed in the subsequent section.

 

 

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