However, it seems this only has complex values.  If we let h = hi where i is the imaginary unit, then we have to solve,

 

6(4+5c+4c²+3e+3ce+3e²)(1+c²+2e+2ce+2e²) = y²

 

or a quartic polynomial in c (or e) that is to be made a square.  One soln is,

 

{c,e} = {(-3-6m+m²)/(-6+2m²),  c-2}

 

and others.  Furthermore, from this rational point e, subsequent ones can then be computed, thus proving eq.1 has an indefinite number of polynomial solns if the terms are complex numbers with rational components.  In summary, for the moment, only particular {a,c,e,g} (extrapolated from Choudhry's cubic) are known such that H₄ has a non-trivial rational h.  If an appropriate general relation can be found between {a,c,e,g} such that H₄ has a quadratic factor with real roots, then it may be possible to find an infinite family of polynomials for k = 1,3,7, just as there is one for k = 1,5.

 

Note:  For k = 1,3,5,7 with ten terms, a similar situation arose, a parametrization dependent on a multi-variable quartic polynomial with only even exponents.  However, this author was able to find two relations between the variables such that the quartic factored.  See discussion in the subsequent sections.

 

 

7.2  Nine terms

 

If x₁ = 0, there are only two known primitive solns to,

 

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k,  for k = 1,3,5,7

 

These lead to ideal solns of deg 8.  Interestingly, terms can be transposed such that x₁+x₂+x₃+x₄+x₅ = y₁+y₂+y₃+y₄+y₅ = 0. (There is a situation similar for k = 1,3,5 wherein the only known solns with x₁ = 0 can be transposed in the same manner.  It begs the question what the first k = 1,3,5,7,9 with x₁ = 0 would look like.)  Letac in 1942  found the two,

 

[0, 34, 58, 82, 98] = [13, 16, 69, 75, 99]

[0, 63, 119, 161, 169] = [8, 50, 132, 148, 174]

 

(How did he find it prior to computer searching?)  (Update, 8/15/09):  Christian Bau did a search for terms < 200 and James Waldby extended it to < 450 with no new primitive solns being found.)  It seems unknown if there is a parametric soln.  But there are two points to be noted.  First, as pointed out by Bailey et al, they can be transposed so an equal number of terms on either side sum to zero.  For the first, it is,

 

[0, -13, -16, -69, 98] = [75, 99, -34, -58, -82]

 

The second one has more "structure" as it can be re-arranged in four distinct ways,

 

[0, 63, 119, -8, -174] = [-161, -169, 148, 50, 132]

[0, 63, 119, -50, -132] = [-161, -169, 148, 8, 174]

 

since 8+174 = 50+132, and,

 

[63, -50, -174, 161, 0] = [-119, 132, 148, -169, 8]

since 169-8 = 161-0.  As the left hand side involves only four terms, this (and only this side) also obeys,

 

5(x₁²+x₂²+x₃²+x₄²)(x₁³+x₂³+x₃³+x₄³) = 6(x₁⁵+x₂⁵+x₃⁵+x₄⁵)

 

a general identity first mentioned in k = 1,3,5.  Second point, for the first soln, re-arranged as,

 

[-99, -13, 0, 34, 98] = [-82, -58, 16, 69, 75]

 

Shuwen found this is good for one even exponent, k = 2.  Later we shall see that both these phenomena appear in the only (and presumably smallest) soln for k = 1,3,5,7,9.

 

Piezas

 

 

This works for the two known solns.  I don’t know whether this is true in general for systems of this form or is just a peculiarity of Letac’s method. 

 

 

7.3 Ten terms

 

The next step is S₇ and S₈,

 

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k

 

 

1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = (-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k

 

 

(3x+2y)^k + (-3x+7y)^k + (3x-2)^k + (3+2x)^k + (-5x-3y)^k =

(-3x+2y)^k + (3x+7y)^k + (-3x-2)^k + (3-2x)^k + (5x-3y)^k

 

for k = 1,2,3,5,7 if x²+6y² = 1 which has the distinction of being valid for k = 2 as well.  Finally, a few others have the condition x²-70y² = 1.  All known parametrizations then for k = 1,3,5,7 in terms of binary quadratic forms involve the unsigned discriminants D = 6, 10, 70.  In short, the set D only employs the small prime factors p = 2,3,5,7.  Whether this is coincidence I do not know.  However, we can show that given the system k = 1,3,5,7,

 

x₁^k+x₂^k+x₃^k+x₄^k+x₅^k = y₁^k+y₂^k+y₃^k+y₄^k+y₅^k

 

if it has either of the constraints,

 

1)      x₁-x₂ = x₃-x₄ = y₁-y₂ 

2)      x₁+x₂ = x₃+x₄ = y₁+y₂  

 

(and equivalent forms after transposition such as the one above), then an infinite sequence of polynomial solns can be found by finding non-trivial rational points on a quartic polynomial that is to be made a square.

 

 

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

 

for k = 2,4, then,

 

a^k + (a+2c)^k + b^k + (b+2c)^k + (-c+d+e)^k = (a+b-c)^k + (a+b+c)^k + d^k + e^k + (3c)^k

 

for k = 1,3,5,7, excluding the trivial case c = 0.  Note that this also satisfies,

 

x₁-x₂ = x₃-x₄ = y₁-y₂ = -2c, 

 

and the additional 2(x₁+x₃) = y₁+y₂ and x₅ = y₃+y₄-(1/3)y₅, which is enough to give the complete soln.   Sinha’s problem then is to solve,

 

p₁^k+p₂^k+p₃^k = q₁^k+q₂^k+q₃^k

 

for k = 2,4 where p₁+p₂-p₃ = 2(q₁+q₂-q₃) and p₁+p₂ ≠ p₃; q₁+q₂ ≠ q₃.  He gave a single polynomial soln to this in terms of binary quadratic forms but this author showed there are four such solns, though there could be more.  

 

1. (-3x+2y+z)^k + (-3x+2y-z)^k + (2x-4y)^k = (8x-y)^k + (2x+3y)^k + (14x-2y)^k

 

where 121x²-10xy-5y² = z²,  D = 70. 

 

2. (5x+2y+z)^k + (5x+2y-z)^k + (-14x-4y)^k = (14x+3y)^k + (4x-y)^k + (6x-2y)^k

 

where x²-50xy-5y² = z²,  D = 70.  (Sinha’s)

 

3. (6x+3y)^k + (4x+9y)^k + (2x-12y)^k = (-x+3y+3z)^k + (-x+3y-3z)^k + (-6x-6y)^k

 

where x²+10y² = z²,  D = -10.  (This one, after minor changes, gives the example at the start of this section.)

 

4. (2x+3y)^k + (-4x+y)^k + (-10x-4y)^k = (3x+y+z)^k + (3x+y-z)^k + (2x-2y)^k

 

where 49x²+40xy+10y² = z²,  D = -10.

 

where D is the discriminant of the binary quadratic form.  Furthermore, treated as rational points on a certain elliptic curve, it can be showed that an infinite sequence can be derived from these thus providing polynomial solns of increasing degree.  (See discussion on Fourth Powers.)  The example below is based on Sinha’s soln, given by (2),

 

(2p+q+r)^k + (-4p-q+r)^k + (5p+q)^k + (9p-q)^k + (11p+3q)^k =

(-2p-q+r)^k + (4p+q+r)^k + (p-2q)^k + (9p+3q)^k + (11p+2q)^k

 

for k = 1,3,5,7,  if p²-50pq-5q² = r².   (If we let {p,q,r} = {x+(25/3)y, (1/3)y, 1}, then this becomes x²-70y² = 1.)

 

Note:  To compare, for eighth powers, other than the Letac-Sinha identity, there are only two solns known found by computer search with all terms x_i < 520.  One obeys the same constraints as Sinha’s identity, x₁-x₂ = x₃-x₄ = y₁-y₂, (and an additional x₄-x₂ = y₄-y₂).  This is,

 

[366, 103, 452, 189, -515] = [508, 245, -18, 331, -471],  for k = 1,2,4,6,8

 

but whether this has a parametric soln is unknown.

Such as this particular one by Gloden,

 

[9, 45, 11, 43, 55] = [3, 51, 19, 37, 53]

 

based on an identity given in the next section and simple transposition of terms,

 

[-51, 9, 11, 55, -53] = [3, -45, -43, 19, 37]

 

gives the equivalent rule,

 

x₁-y₁ = -(x₂-y₂) = -(x₃-y₃)

 

{-71, 143, -17, -163, 121} = {103, -31, 157, -47, -169}

 

which enable me to find a polynomial identity.  This eqn has a lot of structure. Expressed in terms of x_i and y_i, then,

 

x₁-y₁ = -(x₂-y₂) = x₃-y₃

 

which, after negating, is the same as Gloden’s.  But it is also the case that x₃²+x₄² = y₃²+y₄².  Using these constraints, we find the identity given earlier,

 

(3x+2y)^k + (-3x+7y)^k + (3x-2)^k + (3+2x)^k + (-5x-3y)^k =

(-3x+2y)^k + (3x+7y)^k + (-3x-2)^k + (3-2x)^k + (5x-3y)^k

 

for k = 1,2,3,5,7, if x²+6y² = 1.  This has the same properties as the numerical example by Shuwen which is just {x,y} = {29/35, 8/35}, though some rational values should be avoided as they are trivial.  However, one can find a more general result,

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