024: Sixth Powers (6 or 8 terms)

 
 

Piezas

 

We can also consider the more general system,

 

na+b+c = nd+e+f         (eq.1)

ak+bk+ck = dk+ek+fk     (eq.2)

 

with k = 2,6 for any n.  To solve this, we can use solns to n = 1 and exploit the fact that sign changes to the terms ai do not affect eq.2.  The form F1 that gave the complete soln for the case k = 2,4 is also useful here,

 

(p+qu)k + (r+su)k + (t+u)k = (p-qu)k + (r-su)k + (t-u)k

 

hence {a,b,c,d,e,f} = {p+qu, r+su, t+u, p-qu, r-su, t-u} and we will solve the system,

 

a+b-c = d+e-f

na+b+c = nd+e+f

a2+b2+c2 = d2+e2+f2

 

giving {t,s,q} = {-pq-rs,  -1-nq,  2/(1-n)}.  Using these, F1 at k = 6 reduces to the form,

 

(Poly1)u2+(Poly2) = 0

 

where Poly1 is linear in {p,r} and Poly2 is cubic given by,

 

Poly1:= (-3+n)(5-2n+n2)p + 4n(1+n2)r

Poly2:= (-3+n)(5-2n+n2)p3 + 2(5+11n-5n2+n3)p2r + (-5-7n-15n2+3n3)pr2 + 4n(1+n2)r3

 

Thus, again, one should solve,

 

y2 = – (Poly1)(Poly2)

 

a situation similar to the previous section.  Trivial solns should be avoided such as y = Poly1 = 0 or r = -p(n-3)/(2n) with the latter, using Fermat’s method to compute a second rational point, just yielding the former.  (Note that the constraint excludes n = 3, though this in fact is solvable using another method.)  There are other non-trivial points though.  For ex, let n = 12, then,

 

y2 = -75(75p+464r)(225p3+458p2r+587pr2+1392r3)

 

and with {p,r} = {-104, 17} this gives,

 

(-127)k + 496k + 393k = 23k + (-479)k + (-432)k

 

From this initial soln, subsequent ones can then be found proving that the system,

 

a+b-c = d+e-f

12a+b+c = 12d+e+f
ak+bk+ck = dk+ek+fk

 

for k = 2,6 has an infinite number of solns.  Later, it will be shown that this approach can be extended to solve the system,

 

ma+b+c+d = me+f+g+h

ak+bk+ck+dk = ek+fk+gk+hk

 

for any m for k = 2,4,6.

 

Note 1:  For what small integer n does this method provide non-trivial solns?  After some experimentation, I found n = 12, 15, 21, 30, 33, 135 for n < 150 though there could be skips since I was focusing only on n = 3m and assuming relatively small initial solns.  Other than n=1, is there an n ≠ 3m with an infinite number of solns?

 

Note 2:  Excluding the case n=1 of which the complete soln is known, are there other general methods for other n?  For example, the one discussed in this section cannot deal with n=3, though there is a polynomial identity for this.

 

Note 3:  Given a parametric soln to k = 1,2,6, it seems one can modify this to solve na+b+c = nd+e+f.  For ex, using Choudhry’s soln to x1k+x2k+x3k = y1k+y2k+y3k, with {xi, yi} as,

 

{x1, x2, x3} = { 2(α+β)m+(α-β+t),  -2αm+(α+β+t),  -2βm-(α+β-t)}

{y1, y2, y3} = {-2(α+β)m+(α-β+t),   2αm+(α+β+t),   2βm-(α+β-t)}

 

discussed previously which is already true for k = 1,2, a linear combination of terms involving only the constants α,β should be found.  Three possibilities are,

 

nx1+x2-x3 = ny1+y2-y3,    x1+nx2-x3 = y1+ny2-y3,    x1-x2+nx3 = y1-y2+ny3,

 

where n = (α-β)/(α+β), n = (α+2β)/α, and n = (α+2β)/β, respectively.  If either α,β = 1, then the last two yield integer n.  For α = 1, Choudhry found β = {7, 10, 16, 67} so using the second formula this gives n = {15, 21, 33, 135}, all n = 3m and consistent with the odd n found by this author.

 
Note 4: If there is an initial soln for a certain n, is there always an infinite number for this n?

 

Note 5: A small variation of the method and which mostly yields the same results is to use different terms of F1 which are to be multiplied by n and have their signs changed.  Again let {a,b,c,d,e,f} = {p+qu, r+su, t+u, p-qu, r-su, t-u} but to solve the system,

 

-a+b+c = -d+e+f

a+b+nc = d+e+nf

ak+bk+ck = dk+ek+fk, k = 2

 

giving,

 

{t,s,q} = {-pq-rs,  -n-q,  (1-n)/2} and F1 at k = 6 again reduces to the form,

 

(Poly11)u2+(Poly12) = 0

 

though now defined diffently,

 

Poly11:= (-3+n)(5-2n+n2)p + (3+n)(5+2n+n2)r

Poly12:= (-3+n)(5-2n+n2)p3 + (25-7n-5n2+3n3)p2r + (-25-7n+5n2+3n3)pr2 + (3+n)(5+2n+n2)r3

 

As before, solve y2 = – (Poly11)(Poly12) with trivial soln {p,r} = {-(n+3), n-3}.  For ex, for comparison, again let n = 12,

 

y2 = – 75(75p+173r)(225p3+881p2r+1159pr2+519r3)

 

with non-trivial soln {p,r} = {-39, 17} which has the same r and yields the same terms as in the first curve.  To compare the two methods, for n = 12, 15, 21, 30, 33, 135, non-trivial points {p,r} are,

 

First method:   {-104, 17}, {-183, 34}, {-53, 17}, {-24, 13}, {-137, 39}, {-229, 106}

Second method{-39, 17},   {-65, 34},  {-24, 17},    {?, ?},     {-50, 39},  {-111, 106}

 

Notice that the value for r is the same.  If the two methods give the same results, why is there apparently no small non-trivial soln for n = 30 using the latter one?  Furthermore, Poly2 and Poly12 as cubic polynomials have different but related discriminants Di being,

 

D1 = (5-2n+n2)(5+2n+n2
D2 = (5-2n+n2)(1+n2)

 

respectively.  Do the two methods then for the same n always give the same results?

 

Q:  Any other forms for equal sums of sixth powers with six terms?

 

 

6.3 Seven terms

 

It is unknown if the unbalanced multi-grade eqn,

 

x1k+x2k+x3k = x4k+x5k+x6k+x7k,   for k = 2,4,6
 
has non-trivial solns.  Q:  Anyone can provide one, preferably parametric?
 
(Update, 2/22/10):  In response to a post in sci.math, Giovanni Resta made a search and found that there are no solns with terms < 845.  James Waldby would later extend the search radius to < 1280 with still no soln.  However, there are a few that are good only for k = 2,6, namely,
 

[73, 58, 41] = [70, 65, 32, 15]
[85, 74, 61] = [87, 71, 56, 26]
[218, 167, 29]  = [224, 107, 102, 65]
[351, 265, 221] = [336, 309, 169, 73]
[493, 335, 65] = [497, 297, 155, 16]

 

Some exhibit interesting side relations between its terms, such as the smallest one with x1-x2 = x7, x1-x3 = x6, though with appropriate substitutions, they are still not enough to reduce it to a parametric form.  (End update.)

 

 

 

6.4 Eight terms

 

Some general identities analogous to the quadratic-quartic system Q1 are:

 

Theorem (Piezas):  Let ak+bk+ck+dk = ek+fk+gk+hk,  for k = 2,4,6. 

 

Define {x,y,z} = {a2+b2+c2+d2,  a4+b4+c4+d4,  a6+b6+c6+d6}.  Then,

 

a8+b8+c8+d8-e8-f8-g8-h8 = -4(a2-e2)(b2-e2)(c2-e2)(d2-e2) = -4(a2b2c2d2-e2f2g2h2)

 

4(a10+b10+c10+d10-e10-f10-g10-h10) = 5(a8+b8+c8+d8-e8-f8-g8-h8)(x)

 

8(a12+b12+c12+d12-e12-f12-g12-h12) = 6(a8+b8+c8+d8-e8-f8-g8-h8)(x2+y)

 

24(a14+b14+c14+d14-e14-f14-g14-h14) = 7(a8+b8+c8+d8-e8-f8-g8-h8)(x3+3xy+2z)

 

The first one is true regardless of whether e is replaced by any of {f,g,h}. The next two can be combined together to an 8-12-10 Identity.  Define n as,

 

a4+b4+c4+d4 = n(a2+b2+c2+d2)2

 

then,

 

25(a8+b8+c8+d8-e8-f8-g8-h8)(a12+b12+c12+d12-e12-f12-g12-h12) = 12(n+1)(a10+b10+c10+d10-e10-f10-g10-h10)2

 

This is analogous to Ramanujan’s 6-10-8 Identity though since Ramanujan was able to use a linear relationship between the terms ai, namely a+b = ±c, to define n as the constant ½, his result was then much neater.  However, work by Euler shows that,

 

a4+b4+c4+d4 = n(a2+b2+c2+d2)2

 

does have a parametric soln for n = ½, given in the section on Fourth Powers, though it is much more complicated than for the case d = 0.

 

Chernick (and Escott, independently)

 

There is the aesthetically-pleasing identity,

 

(-u+7w)k + (u-2v+w)k + (-3u-w)k + (3u+2v+w)k =

(u+7w)k + (-u+2v+w)k + (3u-w)k + (-3u-2v+w)k

 

for k = 1,2,4,6, if u2+uv+v2 = 7w2

 

with signs changed by this author so it would be valid for k = 1.  This was derived using the equally simple quartic identity,

 

uk + vk + (u+v)k = wk + (2w)k + (3w)k 

 

for k = 2,4 with the same condition.  Since the eqn is homogeneous, one can do the substitution {u,v,w} = {y+z, y-z, x} and the condition becomes,

 

7x2-3y2 = z2

 

where the binary quadratic form has square-free discriminant D = 21.  This is the only known quadratic identity (in contrast to the system k = 1,3,5,7 which has three known D).  So how did Chernick and Escott find this?  One technique is to keep applying the transformation theorems discussed in Chap II to gradually increase the range of the exponents and, if the chosen starting identity has special properties, the same terms may appear on both sides and cancel out, keeping down the number of terms.  For example, using the quartic identity and Theorem 5 with T = 0, this becomes,

 

[u, v, u+v, -u, -v, -u-v] = [w, 2w, 3w, -w, -2w, -3w],  k = 1,2,..5 

 

If the Tarry-Escott theorem is then used, generally this doubles the number of terms.  But for this particular instance, using T = w, the number of terms on either side of the equation instead of doubling, increases only up to eight but the range is now k = 1,2,3,4,5,6.  A second application of the same theorem, with T = u, does not increase the number of terms at all and produces a symmetric ideal soln for k = 1,2,..7.  Since this has appropriate pairs with a common sum, using Theorem 5 in reverse it is easy to recover a (k.4.4) for k = 2,4,6 and which is the identity given at the start of this section.

 

Using a similar method, Tarry earlier found the first ideal soln (symmetric) for k =1,2,..7.  Starting with [1, 9] = [4, 6] which is only for k = 1, he used five successive applications of the Tarry-Escott theorem with T = 2, 1, 7, 8, 13, respectively, to get,

 

[1, 5, 10, 16, 27, 28, 38, 39] = [2, 3, 13, 14, 25, 31, 36, 40],  k = 1,2,…6 

 

Note that this is a symmetric soln with the common sum 41, implying this is dependent on an identity valid for k = 1,3,5.  Then using a last transformation with T = 11 found the ideal soln,

 

[1, 5, 10, 24, 28, 42, 47, 51] = [2, 3, 12, 21, 31, 40, 49, 50],  k = 1,2,…7

 

It turns out this can be generalized with solns depending on the nice eqn x2+7y2 = 8z2.

 

 

Piezas

 

To simplify matters we will start with the sixth deg.  Rearranging terms, this is

 

[5, 16, 27, 38, 28, 39, 1, 10] = [3, 14, 25, 36, 2, 13, 40, 31],  k = 1,2,…6

 

which makes it easy to see that the first six terms (with a skip after the fourth) of each side involve an arithmetic progression, differing by 11.  This is then an example of a “special” property.  To recall, by Frolov’s theorem we can always set a1 = 0 so, to generalize Tarry's result, we are to solve the system,

 

[0, a, 2a, 3a, a+c, 2a+c, x1, x2] = [b, a+b, 2a+b, 3a+b, a+b-c, 2a+b-c, y1, y2],  k = 1,2,…6

 

with unknowns {a,b,c,x1,x2,y1,y2}.  Using the Tarry-Escott theorem, with T = a the properties of the system ensure there is no increase in the number of terms and this becomes,

 

[0, a+c, x1, x2, 4a+b, 3a+b-c, a+y1, a+y2] = [b, a+b-c, y1, y2, 4a, 3a+c, a+x1, a+x2]

 

now for k = 1,2,…7.  But since we wish this to be symmetric, pairs must have a common sum,

 

0 + (4a+b) = (a+c) + (3a+b-c) = (x1) + (a+y1) = (x2) + (a+y2)

b + (4a) = (a+b-c) + (3a+c) = (y1) + (a+x1) = (y2) + (a+x2)

 

where the first two pairs have the sum 4a+b while the last two imply x1+y1 = x2+y2.  (In Tarry’s example, this is 1+40 = 10+31.)  This can be satisfied by letting {x1,x2,y1,y2} = {t+u, t+v, t-u, t-v} and, if these are to have the same common sum as the others, then it must be the case that,

 

(t+u) + (a+(t-u)) = 4a+b

 

so t = (3a+b)/2.  We have reduced our unknowns to just five: {a,b,c,u,v}.  (In fact, it can be assumed c = 1 without loss of generality.)  Using Theorem 5 in reverse, if we subtract (4a+b)/2 from the first four terms on each side, namely {0, a+c, x1, x2} and {b, a+b-c, y1, y2}, this gives us,

 

(4a-b)k + (2a-b+2c)k + (a+2u)k + (a+2v)k = (4a+b)k + (2a+b-2c)k + (a-2u)k + (a-2v)k

 

for k = 2,4,6.  To simplify, let {a,b,c} = {2p, -2q, -q+r} and rearranging terms, we get,

 

(p+u)k + (p+v)k + (2p+r)k + (4p+q)k  = (p-u)k + (p-v)k + (2p-r)k + (4p-q)k,           (eq.1)

 

Using this simpler system, we can then solve for the unknowns.  This is given by,

 

{p,q,r,u,v} = {2x2+x+1,  x2+6x+1,  -8(x2+2x),  6x2+4x-2+y,  6x2+4x-2-y}

 

where x,y must satisfy,

 

y2 = 15x4+104x3+223x2+54x+4

 

with integral solns x = -3,-2,0,1,5 which are trivial though a non-trivial one is x = -1/7.  From these one can then compute more rational points. This was derived by expanding eq.1 at k = 2,4,6 and eliminating {u,v} to get the first condition 8p2-8q2-5qr-r2 = 0.  Letting {p,q,r} = {z, 2y, x-5y}, this becomes,

 

x2+7y2 = 8z2

 

with initial soln {x,y,z}= {1,1,1} and is easily solved parametrically.  By recovering {u,v} there is a second quadratic condition to fulfill hence the elliptic curve.  As was alluded to earlier, this symmetric sixth degree implied a fifth degree identity.  Thus, in summary, if,

 

(3p+q)k + (p+q)k + (p+r)k + (u)k = (3p-q)k + (p-q)k + (p-r)k + (-v)kfor k = 1,3,5, then,

 

(p-u)k + (p-v)k + (2p-r)k + (-4p+q)k  = (-p-u)k + (-p-v)k + (-2p-r)k + (4p+q)kfor k = 1,2,4,6. 
 
Note that the second system satisfies x1-y1 = x2-y2 =  (x3-y3)/2 = -(x4-y4)/4 and a minor change in signs made it valid for k = 1 as well.  From this one can find an ideal seventh degree soln such as the beautifully concise,

 

qk + (2p+r)k + (3p+u)k + (3p+v)k + (5p-u)k + (5p-v)k + (6p-r)k + (8p-q)k =

(-q)k + (2p-r)k + (3p-u)k + (3p-v)k + (5p+u)k + (5p+v)k + (6p+r)k + (8p+q)k

 

for k = 1,2,…7 where {p,q,r,u,v} are as defined above.  (Update, 8/4/09):  Alternatively, the system can be expressed in a more symmetric form,
 

Theorem:  If  (a-b-c)k + (a-b+c)k + (b+3c)k + (b+c)k = (a+b-d)k + (a+b+d)k + (-b+3c)k + (-b+c)k,  for k = 1,3,5, then,

 

(a+b+c+d)k + (a+b+c-d)k + (-a+b+2c)k + (b-4c)k  = (a+b-c+d)k + (a+b-c-d)k + (-a+b-2c)k + (b+4c)k,  for k = 1,2,4,6, 
 

where {a,b,c,d} satisfy a2+3ab+4b2 = 8c2 and a2-5ab+36b2 = 8d2.  Note how in the second system, the LHS difers from the RHS only in the sign of the variable c.  This identity is easily proven by solving for {c,d} in terms of {a,b} and subsituting into the two systems, though the ratio a/b = {-4, -2, -1, 1, 4} should be avoided as they are trivial, with a small non-trivial soln as {a,b} = {14,1}. (End note.)

 

Note: Eq.1 can be generalized to the form,

 

(an+p)k + (bn+q)k + (n+r)k + (n+s)k = (an-p)k + (bn-q)k + (n-r)k + (n-s)k

 

for some constants a,b.  Another soln can be found when {a,b} = {1,7}.  Deriving the other variables we get {r,s} = {-p-4q,  -3q} where p,q,n satisfy p2+4pq+7q2 = 28n2.  Let {p,q,n} = {2x-y, y, z} and this becomes,

 

x2+xy+y2 = 7z2

 

and an inspection reveals this is just the Chernick-Escott identity in disguise.  I am not aware of another a,b such that the form has a non-trivial soln though there may be more. The next example also involves a quartic elliptic curve.
 
 
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