14. Form: a^k+b^k+c^k = d^k+e^k+f^k  (k = 2,3,4)

 

A. Choudhry

 

(ay+b)^k + (cy+ad)^k + (-ay+b+cd)^k = (ay-b)^k + (cy-ad)^k + (-ay-b-cd)^k

 

This is already valid for all variables when k=2.  But let,

 

{a,b,c,d} = {x²+3,  (x+3)³(x-1),  -2(x²-3),  2(x²+2x+3)}, 

 

where x satisfies the elliptic curve,

 

y² = (x²-4x+3)²-48x²

 

and this also becomes true for k = 3,4.  Some integer points on this curve are given by x = {-3, -1, 0, 15} while one non-trivial soln is x = 1/5, and so on.

 

Piezas

 

Rational solns to a^k+b^k+c^k = d^k+e^k+f^k,  k = 2,3,4 imply a rational soln t to the quadratic eqn,

 

(a²+b²-e²-f²)t² - 2(a³+b³-e³-f³)t + (a⁴+b⁴-e⁴-f⁴) = 0

 

Proof: The above is equivalent to the identity,

 

(c²-d²)t² - 2(c³-d³)t + (c⁴-d⁴) = 0,  where t = c+d.

 

 

15. Form: x⁴+y⁴+z⁴ = 2(x²y²+x²z²+y²z²)-t²

 

This is equivalent to (x+y-z)(x-y+z)(-x+y+z)(x+y+z) = t² hence is essentially the problem of finding a triangle with rational area using Heron’s formula.  One approach is to reduce it to the equation p²+q²+r² = s² of which the complete soln by Desboves is known.  Equate the first three factors as,

 

{(x+y-z), (x-y+z), (-x+y+z)} = {p², q², r²},

 

and by solving for x,y,z, and substituting into the last factor as x+y+z = s², one then gets the afore-mentioned condition.  Other ways are,

 

1. Brahmagupta

 

{x,y,z} = {a²/b+b,  a²/c+c,  a²/b+a²/c-b-c}

 

2. Kurushima (Euler would later give a similar identity.)

 

{x,y,z} = {ac(b²+d²),  bd(a²+c²),  (bc+ad)(ab-cd) }

 

3. Gauss

 

{x,y,z} = {abcd(a²+b²),  ab(c+d)(a²c-b²d),  ab(a²c²+b²d²) }

 

4. A. Martin

 

Let {m,n} = {a²+b², a²-b²}, then,

 

{x,y,z} = {(c²-d²)m,  2cdm+2d²n,  (c²+d²)m+2cdn}

 

 

16. Form: v⁴+x⁴+y⁴+z⁴ = nt^k

 

There is yet no polynomial identity for a⁴+b⁴+c⁴+d⁴ = t⁴ so this form is the closest so far.

 

E. Fauquembergue

 

(2a²bc³)⁴ + (2ab²c³)⁴ + (2ab(a⁴+b⁴))⁴ + ((a²-b²)c⁴)⁴  = (4a²b²(a⁴+b⁴)²-c¹²)²,  if a²+b² = c²

 

Any other polynomial soln to a⁴+b⁴+c⁴+d⁴ = nt^k?

 

(Note: Yes, there is for n = 18, 63, etc. Add the ones by Haldemann.)

 

L. Jacobi, D. Madden

 

They recently found (in 2008) a general soln to the case a⁴+b⁴+c⁴+d⁴ = (a+b+c+d)⁴ in terms of an elliptic curve.  (I've gone through their work and will post the details when I can.)

 

 

17. Form: v^k+x^k+y^k+z^k = a^k+b^k+c^k+d^k,  k = 2,4

 

Piezas

 

Theorem: If a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k,  for k = 2,4, and abcd = efgh.  Define n as a⁴+b⁴+c⁴+d⁴ = n(a²+b²+c²+d²)² then,

 

32(a⁶+b⁶+c⁶+d⁶-e⁶-f⁶-g⁶-h⁶)(a¹⁰+b¹⁰+c¹⁰+d¹⁰-e¹⁰-f¹⁰-g¹⁰-h¹⁰) = 15(n+1)(a⁸+b⁸+c⁸+d⁸-e⁸-f⁸-g⁸-h⁸)²

 

Note if d = h = 0, the condition abcd = efgh disappears and this reduces to the theorem given earlier of which Ramanujan’s 6-10-8 Identity is a special case.  Proof:  Given,

 

a²+b²+c²+d² = e²+f²+g²+h²       (eq.1)

a⁴+b⁴+c⁴+d⁴ = e⁴+f⁴+g⁴+h⁴       (eq.2)

 

Eliminate h between the two to get a quadratic in {a,b,c,d} and solve for any, say d.  Solve for h in eq.1 and we get the complete radical soln of eq.1 and eq.2.  Substitute these two values into the 6-10-8 identity and it has a factor, call this P, which is the same polynomial that results when the values are substituted into abcd-efgh = 0.  Thus if the latter is satisfied so is the former. (End proof.)   A solution will be given below.

 

Theorem. If a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k,  for k = 2,4,6, where abcd = efgh, and a+b ≠ ±(c+d); e+f ≠ ±(g+h), (call the entire system V₁) then,

 

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

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

 

for k = 1,2,4,6,8,10.

 

Again, for d = h = 0, the condition abcd = efgh disappears so reduces to Birck’s Theorem (given later). This is then an extension to tenth powers and the proof will be discussed in that section.  Even for non-zero variables it is possible one or two of the addends will vanish, as we shall see later.  This author  hasn’t been able to find a parametric soln to V₁ but was able to do so for the simpler case when valid only for k = 2,4.  To solve V₁ (though not completely) including the side conditions, given,

 

x₁^k + x₂^k + x₃^k + x₄^k = x₅^k + x₆^k + x₇^k + x₈^k,  k = 2,4

 

assume it to have the form, call it F₁,

 

a^k(p+q)^k + b^k(r-s)^k + c^k(r+s)^k + d^k(p-q)^k = a^k(p-q)^k + b^k(r+s)^k + c^k(r-s)^k + d^k(p+q)^k

 

One can see that x₁x₄ = x₅x₈ and x₂x₃ = x₆x₇ which takes care of the requirement x₁x₂x₃x₄ = x₅x₆x₇x₈.  To simplify matters, assume further that x₁+x₂ = x₅+x₆.  After some algebra, we get the soln,

 

{p,q,r,s} = {ab²-ac²,  be,  a²b-bd²,  ae},  where a²(3b²-c²) + e² = (b²+c²)d².

 

This conditional eqn is easily solved parametrically.  First, let the LHS be a²n+e² = y² and find {a,e}, then the RHS as b²+c² = z² and find {b,c}, to get the form y² = z²d² which gives d.  One special case satisfies the additional relationship x₁-x₂+x₃+x₄ = -x₅+x₆+x₇+x₈ = 0 and the significance is that, when used on the theorem, one term on each side of the equation vanishes thus reducing the number of terms.  An explicit soln is then,

 

(p+q)^k + a^k(r-s)^k + (r+s)^k + b^k(p-q)^k = (p-q)^k + a^k(r+s)^k + (r-s)^k + b^k(p+q)^k,  k = 2,4

 

where {b,p,q,r,s} = {3a,  -1+a²,  a+3a³,  a-9a³,  1+3a²}.  A numerical example is,

 

83^k + 316^k + 123^k + 276^k = 69^k + 164^k + 237^k + 332^k

 

(Note:  If the particular form F₁ is extended to k = 6, it yields only trivial solns so another form has to be used.)

 

 

18. Form: 2(v⁴+x⁴+y⁴+z⁴) = (v²+x²+y²+z²)²

 

The equation,

 

2(a⁴+b⁴+c⁴) = (a²+b²+c²)²

 

has the simple soln a+b = c, which is central to Ramanujan's 6-10-8 Identity.  Turns out when there are four terms,

 

2(v⁴+x⁴+y⁴+z⁴) = (v²+x²+y²+z²)²    (eq.1)

 

it has a parametric soln as well.  Note that this is a special case of Descartes' Circle Theorem,

 

2(x₁²+x₂²+x₃²+x₄²) = (x₁+x₂+x₃+x₄)²

 

where the x_i are squares.  Euler’s clever soln uses the fact that eq.1 is equivalent to either,

 

(2xy)² + (2vz)² = (-v²+x²+y²-z²)²

(2xz)² + (2vy)² = (-v²+x²-y²+z²)²

 

so is a special case of two simultaneous Pythagorean triples.  More generally, for any d, we are to solve,

 

2(d²v⁴+x⁴+y⁴+z⁴) = (dv²+x²+y²+z²)²   (eq.2)

 

and Euler's soln can be given as {v,x,y,z} = {2pqrs(a²-t²),  t(a²-t²),  t(abrs-cpqt),  a(abrs-cpqt)}, where {a,b,c} = {pq(r²-ds²),  p²+dq²,  r²+ds²}, and {p,q,r,s,t} satisfies the conditional eqn,  

 

pqrs(p²-dq²)(r²-ds²) = t²

 

For d = 1, one soln is {p,q,r,s} = {2m²-n²,  m²+n²,  m²+n²,  m²-2n²}.  Note:  Strictly speaking, Euler’s analysis was focused on the case d = ±1 and this author inserted the d to generalize it.  Any soln for other d?

 

(Update, 9/13/09):  I noticed we can have a better form if we assume t = rs(r²-ds²) and the conditional eqn becomes the symmetric,

 

pq(p²-dq²) = rs(r²-ds²)

 

This has appeared already in Section 013 and has many solns and is connected to the two distinct eqns,

 

a⁴+b⁴ = c⁴+d⁴

a^k+b^k+c^k = d^k+e^k+f^k,  k = 2,4

 

Thus, we can also solve 2(d²v⁴+x⁴+y⁴+z⁴) = (dv²+x²+y²+z²)² as:

 

I.  When d = -1:

 

{v,x,y,z} = {2(ab-cd)(ab+cd),  (a²+b²+c²+d²)(a²-b²+c²-d²),  2(ac-bd)(a²+c²),  2(ac-bd)(b²+d²)},  where a⁴+b⁴ = c⁴+d⁴.

 

II.  When d = 1:

 

Let a^k+b^k+c^k = d^k+e^k+f^k,  for k = 2,4, where a²+b² = f²,  d²+e² = c²,  and ab = de, then,

 

{v,x,y,z} = {ae-bd,  (a+d)(a-d),  a(c+f),  d(c+f)}

 

A soln given by Gloden is {a,b,c,d,e,f} = {p²-q², p²+r², 2pq, p²+q², p²-r², 2pr},  where p² = q²+qr+r².  This can be solved by {p,q,r} = {u²+uv+v²,  u²-v², 2uv+v²}.  (End update.)

 

 

19. Form: x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k,  k = 1,2,3,4

 

The system of eqns,

 

x₁^k+x₁^k+…+ x_n^k = y₁^k+y₁^k+…+ y_n^k,  k = 1,2,3,4

 

has non-trivial solns only for n > 4.  However, it is possible two terms on one side are equal to zero.

 

Piezas

 

Given a^k+b^k+c^k+d^k+e^k = f^k+g^k+h^k,

 

for the special case when h = a+b+c, then,

 

{d,e,f,g} = {(p+u)/(2q),  (p-u)/(2q),  (p+v)/(2q),  (p-v)/(2q)}

 

where p = (a+b)(a+c)(b+c),  q = (ab+ac+bc), and {a,b,c} satisfies the two quadratic conditions,

p(p+4abc) = u²

p(p+4abc) - 4q³ = v²

 

An example of a particular soln is {a,b,c} = {-21, -4, 12} which yields the eqn,

 

(-63)^k + (-12)^k + 36^k + (-35)^k + 10^k = (-62)^k + 37^k + (-39)^k

 

Q: Any general polynomial identity for this?

 

 

20. Form: x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k,  k = 1,2,3,4

 

The complete soln to this system, call this E₄, is unknown.  However, for the case when x₂+x₃ = y₂+y₃ it can be shown that this involves making a quartic polynomial a square.  Use the form,

 

(a+bj)^k+(c+dj)^k+(e+fj)^k+(g+hj)^k+(i+j)^k = (a-bj)^k+(c-dj)^k+(e-fj)^k+(g-hj)^k+(i-j)^k

 

Let f = -d to satisfy the constraint and i = -(ab+cd+ef+gh), h = -(1+b+d+f) for k = 1,2.  Expand this for k = 3,4, to get the auxiliary resultants,

 

(Poly10)j²+(Poly20) = 0                                   (eq.1)

(Poly11)j⁴+(Poly21)j²+(Poly31) = 0                 (eq.2)

 

where the Poly_i are in {a,b,c,d,e,g}.  Eliminate j between eq.1 and 2 (set j = √v for convenience) and one gets a final resultant, call this R, which is only a quadratic in a,c,e,g.  Solving for g, this has a discriminant D which is only a quadratic in e so to solve D = y², one uses a quadratic form e = Q(x) for some variable x.  After a rational g is found, to find j, solve,

 

-(Poly10)(Poly20) = z²

 

which is a quadratic in e. Substituting the quadratic form e = Q(x) into this, it becomes,

 

Poly(x) = z²          (eq.3)

 

which is a quartic in x, hence any soln to E₄ with the constraint x₂+x₃ = y₂+y₃ must satisfy eq.3, a quartic polynomial to be made a square.

 

Note:  Since the final resultant R involves six variables {a,b,c,d,e,g}, the explicit soln is a mess.  There might be a way to clean this up though using a more suitable form than the one given without losing its generality.  But using the simpler form,

 

(a+b)^k+(c+d)^k+(e+f)^k+(g+h)^k+(i+j)^k = (a-b)^k+(c-d)^k+(e-f)^k+(g-h)^k+(i-j)^k

 

for example, is messier.

 

 

21. Form: x₁⁴+x₂⁴+…x_n⁴, n > 4

 

Most of the identities here are for five fourth powers equal to a fourth power.

 

E. Fauquembergue

 

(4x⁴-y⁴)⁴ + 2(4x³y)⁴ + 2(2xy³)⁴ = (4x⁴+y⁴)⁴

 

C.Haldeman

 

(2x²+12xy-6y²)⁴ + (2x²-12xy-6y²)⁴ + (4x²-12y²)⁴ + (3x²+9y²)⁴ + (4x²+12y²)⁴ = 5⁴(x²+3y²)⁴

 

This gives as a first instance 2⁴+2⁴+4⁴+3⁴+4⁴ = 5⁴.  Similar identities were found by A. Martin and Ramanujan.  A generalization has been found by this author.

 

Piezas

 

(ax²+2v₁xy-3ay²)^k + (bx²-2v₂xy-3by²)^k + (cx²-2v₃xy-3cy²)^k = (a^k+b^k+c^k)(x²+3y²)^k

 

where {v₁, v₂, v₃, c} = {a+2b, 2a+b, a-b, a+b}, for k = 2,4

 

Thus, it suffices to decompose a sum which starts as a⁴+b⁴+(a+b)⁴+… into a sum and difference of any number of biquadrates and apply it to the identity.  For example, using the eqn 50⁴+50⁴+100⁴+4⁴+15⁴ = 103⁴, this yields, 

 

(50x²+300xy-150y²)⁴ + (50x²-300xy-150y²)⁴ + (100x²-300y²)⁴ + (4x²+12y²)⁴ + (15x²+45y²)⁴ = 103⁴(x²+3y²)⁴

 

R. Norrie

 

(x+y)⁴ + (-x+y)⁴ + (2y)⁴ + (u²-v²)⁴ + (2uv)⁴ = (u²+v²)⁴,  if  2uv(u²-v²) = x²+3y²

 

Note that {u²-v², 2uv} are the legs of a Pythagorean triple.  Thus, if the product of the legs can be expressed as x²+3y², it then gives a soln to the above equation, the smallest of which is the particular equation given earlier, {2,2,4,3,4; 5}.  For the sum of six biquadrates,

 

R.Carmichael

 

(a⁴-2b⁴)⁴ + (2a³b)⁴ + 4(2ab³)⁴ = (a⁴+2b⁴)⁴

 

Note that the form (a⁴+2b⁴)⁴ can also be expressed as a sum of both squares and biquadrates,

 

E. Fauquembergue

 

(a⁴-2b⁴)⁴ + (2a³b)⁴ + (8a²b⁶)² = (a⁴+2b⁴)⁴

 

(2a²b²)⁴ + (2a³b)⁴ + (a⁸-4a⁴b⁴-4b⁸)² = (a⁴+2b⁴)⁴

 

 

 

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