028: Eighth Powers

 
 

PART 11.  Sum / Sums of Eighth Powers 

 

8.0 Six terms

8.1 Eight terms

8.2 Nine terms

8.3 Ten terms

8.4 Twelve terms

8.5 Fourteen terms

8.6 Sixteen terms

8.7 Eighteen terms

 

 

8.0  Six terms 

 

It is conjectured that,

 

x18 + x28 + x38 = y18 + y28 + y38

 

is not solvable other than trivial permutations of the xi yi.  However, it can be shown that there is an infinite number of solutions to,

 

x18 + x28 + x34 = y18 + y28 + y34

 

Proof:

 

We simply use the identity,

 

(a+b)8 + (a-b)8 + (4ab)4 = 2(a4+14a2b2+b4)2

 

(Note: The RHS is in fact a polynomial invariant of the tetrahedron.) Thus, to solve,

 

(a+b)8 + (a-b)8 + (4ab)4 = (c+d)8 + (c-d)8 + (4cd)4

 

it suffices to deal with,

 

a4+14a2b2+b4 = c4+14c2d2+d4

 

Choudhry has given a 7th-deg parametrization (see also Form 3 here) to,

 

 x14+2nx12x22+x24 = y14+2ny12y22+y24

 

as, 

 

{x1, x2} = {1+n2t+st2-rt3+qt4+3t5+pt6+nt7,  -n+pt-3t2+qt3+rt4+st5-n2t6+t7}

{y1, y2} = {1-n2t+st2+rt3+qt4-3t5+pt6-nt7,    n+pt+3t2+qt3-rt4+st5+n2t6+t7}

 

where {p,q,r,s} = {1-n+n2,  -2+4n+n2,  -4n+n3, 1+n2+n3} for free variable t.  For n = 7, and starting with t = 2, we get after removing common factors,

 

11188 + 19378 + 25020454 = 4558 + 18468 + 32006914

 

and so on for other values of t.

  

 

8.1  Eight terms 

 

The first soln to,

 

x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k,  k = 8

 

was found by Nuutti Kuosa in 2006,

 

8618 + 19538 + 20128 + 31138 = 11288 + 25578 + 27678 + 28238

 

though it is not known if this the smallest possible as this is the only soln so far.  It is also not known if k = 1,8 or k = 1,2,8 is possible.  Based on known solns to k = 1,3,5, k = 1,2,6, and k = 1,3,7, this author is assuming a k = 1,8 will be found that satisfies the side conditions: x1+x2 = y1+y2, and x3+x4 = y3+y4.

 

 

8.2  Nine terms

 

The first and only soln so far to 8 eighth powers equal to an eighth power,

 

x18+x28+x38+x48+x58+x68+x78+x8 = z8

 

was found by Scott Chase in 2000,

 

908 + 2238 + 4788 + 5248 + 7488 + 10888 + 11908 + 13248 =  14098

 

Whether there are seven 8th powers equal to an 8th power is unknown, though they are conjectured to be possible (the Lander, Parkin, and Selfridge Conjecture in Eulernet).  No more k positive kth powers equal to a kth power is known for k > 8 (with the case k = 6 unknown as well), though J. Wroblewski has come close for k = 9,10.

 

 

 8.3   Ten terms

 

Other than particular instances of the Letac-Sinha identity, only two solns are known for multi-grade k = 2,4,6,8 (found by Borwein, Lisonek, and Percival in 2000) namely,

 

[71, -131, -180, 307, 308] = [301, 99, 100, 188, -313]     (eq.1)

[366, 103, 452, 189, -515] = [508, 245, -18, 331, -471]   (eq.2)

 

which, as signed terms, are also valid for k = 1 hence all known solns are for k = 1,2,4,6,8.  (Note:  After all, there are 32 possible sums of   ± x1 ± x2 ± x3 ± x4 ± x5 and an equal number for the other side, and the chances might be that two of the sums will be the same.  In general, it raises the question whether systems k = 2,4,…2n as n increases also tend to be valid for k = 1 if the terms are signed.) 

 

Eq.1 has two side conditions: x1-x2 = y1-y2 and x4-x5 = y2-y3.  Eq. 2 is more structured, satisfying five:  x1-x2 = x3-x4 = y1-y2 = y2-y3, and x1-x4 = y1-y4, and x2-x4 = y2-y4.  It is easily shown that, together with the five eqns k = 1,2,4,6,8, (eq. 2) is the only non-trivial soln of this system of 10 eqns in 10 unknowns.

 

A. Letac, T. Sinha 
 
(a+c)k + (a-c)k + (3b+d)k + (3b-d)k + (4a)k = (3a+c)k + (3a-c)k + (b+d)k + (b-d)k + (4b)k

 

for k = 1,2,4,6,8, where a2+12b2 = c2  (eq.1), and 12a2+b2 = d2  (eq.2).  The ratio a/b = {1/2, 2} must be avoided as it yields trivial solns, with the smallest non-trivial one as {a,b} = {218, 11869}.  Labeled as,

 

x1k+x2k+x3k+x4k+x5k = y1k+y2k+y3k+y4k+y5k

 

this also satisfies x1-x2 = y1-y2, and x3-x4 = y3-y4, and interestingly, for k = 1,2, 

 

x1k+x2k+x5k = y1k+y2k;    x3k+x4k = y3k+y4k+y5k

 

for any {a,b,c,d}.  (In fact, it can be shown that the complete soln to the system with all these constraints is given by this identity.)  The first pair of constraints implies that the system k = 2,4,6,8 may have the “side condition”,

 

x1 ± x2 = y1 ± y2;   x3 ± x4 = y3 ± y4

 

just like for sixth powers k = 2,4,6.  To solve the Letac-Sinha identity, since the complete soln to eq.1 is given by {a,b} = {m2-12n2, 2mn}, if we substitute this into eq.2 we get,

 

4(3m4-71m2+432) = d2

 

where it was set n=1 without loss of generality.  Trivial rational solns are given by m = {2, 3, 4, 6}.  However, we can use these to generate subsequent ones that are non-trivial, such as m = 109, and so on.

 

 

(Update, 2/1/10):  Chris Smyth 

 

This is a variation of the Letac-Sinha 8th power [k.5.5] multi-grade.  For k = 1,2,4,6,8,

 

[xy+ax+ay-d, -(xy-ax-ay-d), xy-bx+by+d, -(xy+bx-by+d), cx+cy]k =

[xy+bx+by-d, -(xy-bx-by-d), xy-ax+ay+d, -(xy+ax-ay+d), -(cx-cy)]k

 

where {a,b,c,d} = {1, 3, 4, 11} and {x,y} satisfies the eqn x2y2-13(x2+y2)+112 = 0. 

 

One can see in the general form how “a” in one side has just been replaced with “b” in the other side.  With terms expressed as {pi,qi}, this obeys,

 

p1-p2 = q1-q2

p3-p4 = q3-q4

 

If we set that,

 

p1k+p2k+p3k+p4k+p5k = q1k+q2k+q3k+q4k+q5k,  k = 1,2

 

then one must have 2ak-2bk+ck = 0,  for k = 1,2, and the consequence is that,

 

p1k+p2k+p5k = q1k+q2k, 

p3k+p4k = q3k+q4k+q5k

 

for k = 1,2 which, naturally enough, is the same set of constraints obeyed by the Letac-Sinha identity.  Expanding the system for k = 4,6,8, it will be seen that the only non-trivial soln in the rationals is {a,b,c,d} = {m, 3m, 4m, 11m2} where one can set m = 1 without loss of generality and {x,y} satisfies the eqn,

 

x2y2-13(x2+y2)+112 = 0

 

Let {x,y} = {u, v/(u2-13)} and this is the elliptic curve,

 

(u2-13)(13u2-112) = v2

 

Smyth’s form for (k.5.5) can be generalized by adding two terms on each side to create a (k.7.7).  A possible addition, given in italics, may be,

 

[xy+ax+ay-c, -(xy-ax-ay-c), xy-bx+by+c, -(xy+bx-by+c), dx+dy, ex+fy, fx-ey]k =

[xy+by+bx-c, -(xy-by-bx-c), xy-ay+ax+c, -(xy+ay-ax+c), dx-dy, ey+fx, ex-fy]k

 

which is the same pair successfully added to the (k.4.4).  For non-zero {e,f}, whether this has a non-trivial soln up to k = 8, 10, or even 12 is unknown.  See also Sixth Powers for the context of this identity. (End update.)

 

 

(Update, 1/13/10):  Wroblewski

 

Using a special case of a general theorem, Wroblewski used a k = 2,4,6,8 to derive a k = 1,3,5,7,9!

 

“If [x1, x2, x3, x4, x5]k = [y1, y2, y3, y4, y5]k,   for k = 2,4,6,8,

 

where x1-x2 = x3-x4 = y1-y2 = y2-y3 = N, then,

 

[a+x5, -a+y4, -a+y3, a+x1, a+x3, -a+y5]k = [-a+x5, a+y4, a+y1, -a+x2, -a+x4, a+y5]k,  for k = 1,3,5,7,9

 

where a = N/2.”

 

It can be seen that the second eqn (or eq. 2) found by Borwein et al,

 

[366, 103, 452, 189, -515] = [508, 245, -18, 331, -471] 

 

satisfies this and yields,

 

[-767, 399, -299, 995, 1167, -1205] = [-1293, 925, 1279, -57, 115, -679] 

 

which is the second k = 1,3,5,7,9 since the first one was found in 2000 by Shuwen via computer search.  Since then, Wroblewski has found a few others also by computer search.  See also Ninth Powers.  (End update.)

 

(Update, 1/27/10)Piezas

 

It can be shown that the system,

 

[x1, x2, x3, x4, x5]k = [y1, y2, y3, y4, y5]k,   for k = 2,4,6,8,

 

with the constraint x1-x2 = x3-x4 = y1-y2 = y2-y3 can be completely solved in terms of nested square roots.  The soln is,

 

(1-e)k + (-1-e)k + (1-f)k + (-1-f)k + ak  = (2-c)k + (-c)k + (-2-c)k + bk + dk  

 

a = 4(u+v),

b = 4(u-v)

2c2 = -3+t+31uv-3w

2d2 = 53+t+71uv-11w

{e2, f2} = {p-q,  p+q}

 

and,

 

p = (1/2)(13+t+9uv-5w)

(1/2)q2 = 20+4t+301uv+35u2v2-5(4+11uv)w

 

where w2 = -15+2t+22uv-135u2v2, and t = 32(u2+v2), for two free variables {u,v}.  Thus {e,f} are two roots of an octic polynomial. This then leads to the 9th deg multigrade,

 

(1+a)k + (-1+b)k + (-3+c)k + (-1+d)k  + (2+e)k + (2+f)k =

(-1+a)k + (1+b)k + (3+c)k + (1+d)k + (-2+e)k + (-2+f)k,   for k = 1,3,5,7,9

 

Six variables S = {c,d,e,f,q,w} must be squares, and ±w is equally valid, though only one sign may yield a rational value.  One non-trivial soln is {u,v} = {46/263,  423/526} which gives the original equation that inspired the form.  Even with six expressions to be made squares, due to the symmetric nature of the system, there is in fact an infinite number of {u,v} such that, with the sign of ±w chosen appropriately, all S are squares, one of which is the family,

 

v = 4(u-1)/(12u-7)

 

but unfortunately just yields trivial terms.  (To find two others, one can expand at k = 11.)  However, it may be possible there are other non-trivial solns.  If we add one more constraint to the 8th deg system, namely x2-x4 = y2-y4, we lose one free variable but the problem simplifies to four quadratics to be made squares. This is given by, 

 

[b-c-d, b+c-d, -b-c-d, -b+c-d,  f]k = [a+b-2c,  a+b, a+b+2c, a-b, e]k,   for k = 2,4,6,8

 

where {c,d,e,f} are,

 

32c2 = 5a2-10ab+8b2

32d2 = 29a2+6ab+72b2

4e2 = 9a2-6ab+36b2

4f2 = 13a2+2ab+4b2

 

Since the eqns are homogeneous, we can set b = 1 without loss of generality and "a" is just the only free variable.  This system has also been investigated by Wroblewski, see below.  There are some trivial {a,b} like the ratio a/b = {2, 6, etc).  A non-trivial soln is {a,b} = {-288, 43} which makes all {c,d,e,f} as squares and yields the original multi-grade.  It is easy to make {c,d} as squares using an elliptic curve, though apparently it is only by chance that {e,f} are squares as well.  But this can provide a rational family to,

 

x1k + x2k + x3k + x4k + x5k/2 = y1k + y2k + y3k + y4k + y5k/2for k = 2,4,6,8

 

Considering how this result has a much simpler form, it may be possible that, with the right constraint, the more general case can be simplified still.  But if we add yet one more condition and require that k = 1 hold as well (for signed terms), then it can be shown that the original eqn is the only non-trivial soln to this system of 9 eqns (4+5) in 9 unknowns (since one variable can be set equal to 1 without loss of generality).  (End update.)

 

Piezas

 

Theorem: If ak+bk+ck+dk+ek = fk+gk+hk+ik+jk, for k = 2,4,6,8, define n as, a4+b4+c4+d4+e4 = n(a2+b2+c2+d2+e2)2,  then,

 

72(a10+b10+c10+d10+e10-f10-g10-h10-i10-j10)(a14+b14+c14+d14+e14-f14-g14-h14-i14-j14) = 35(n+1)(a12+b12+c12+d12+e12-f12-g12-h12-i12-j12)2

 

which is a part of a family that starts with Ramanujan’s 6-10-8 Identity and depends on a system that is valid for k = 2,4,…2n. 

 

Note 1:  It can be shown that the system, (pa+c)k + (pa-c)k + (qb+d)k + (qb-d)k +(ra)k = (qa+c)k + (qa-c)k + (pb+d)k + (pb-d)k +(rb)k,  for k = 1,2,4,6,8, where c = √(ma2+nb2) and d = √(na2+mb2) is non-trivial only for {p,q,r} = {1, 3, 4}, and {m,n} = {1, 12}.

 

Note 2:  Just like for k = 1,3,5,7, the general k = 1,2,4,6,8 can be completely parametricized by 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

 

where for simplicity set a = 1 without loss of generality.  For the system which also has x1-x2 = y1-y2,  x3-x4 = y3-y4 it can be shown this entails the necessary but not sufficient condition of finding a non-trivial rational root of a 10th deg resultant.  Let d = b, h = f.  To satisfy k = 1,2, one can then use the variables i,f.  Expanding for k = 4,6,8 gives the eqns,

 

(Poly01)j2 + (Poly02) = 0                                                         (eq.1)

(Poly11)j4 + (Poly12)j2 + (Poly13) = 0                                     (eq.2)

(Poly21)j6 + (Poly22)j4 + (Poly23)j2 + (Poly24) = 0                 (eq.3)

 

Eliminate j between eqs.1,2, then between eq.1,3, (as before, let j = √x for ease of computation) to get two auxiliary resultants, eq.4a and eq.5a, which are 6th and 9th deg in the variable g, respectively. There are still 4 variables, b,c,e,g, and it is at this point that the system for k = 1,2,4,6 with a similar “side condition” has a resultant with a single non-trivial linear factor and some trivial ones.  For k = 1,2,4,6,8, one more variable has to be eliminated but since eq.4a and 5a are of high degree, these are horrendous to resolve directly.  A trick to reduce the degree (which works for this and similar constraints) is to let {e,g} = {u+v, u-v} to get,

 

(Poly31)v4 + (Poly32)v2 + (Poly33) = 0                                    (eq.4b)

(Poly41)v6 + (Poly42)v4 + (Poly43)v2 + (Poly44) = 0               (eq.5b)

 

where again let v = √y to shorten the calculation.  Eliminate y and the final resultant turns out to have a non-trivial 10th-deg factor in c, with trivial linear ones, showing a qualitative difference from the previous system which had a non-trivial linear factor.  One must then find appropriate {b,u} such that this decic has a non-trivial rational root and the Letac-Sinha identity guarantees there are infinitely many.  Whether there is another family such that this decic has rational factors is not known.  In summary, the system,

 

x1k+x2k+x3k+x4k+x5k = y1k+y2k+y3k+y4k+y5k

 

as S7 for k = 1,3,5,7 and S8 for k = 2,4,6,8 can have either of the side-conditions, 

1)      x1-x2 = x3-x4 = y1-y2

2)      x1+x2 = x3+x4 = y1+y2 (or equivalent forms)

3)      x1-x2 = y1-y2;  x3±x4 = y3±y4 

 
with S7 having solns for all three and S8 only for (1), (3).  Whether all three can apply to both remains to be seen.  Furthermore, for powers greater than the sixth, it seems more constraints are needed to decrease the number of variables so that: a) a parametric soln is possible; or b) the final resultant is not of so high degree. 
 
For example, the Letac-Sinha identity which is the only one known for eighth powers already satisfies four or five side conditions.  This implies that for the 12-term nonic and decic systems, and analogous higher ones, a lot of algebraic ingenuity must be employed to find a parametrization.  If there is any.  Like solving univariate eqns in the radicals, there might come a point when there are simply too many variables than can be handled by the constraints.  (See update in Tenth Powers.)
 
 
(Update, 1/12/10):  Piezas, Wroblewski

 

Starting with the known soln for k = 2,4,6,8,

 

[-508, -245, 331, 18, 471] = [-366, 189, -103, 452, 515]

 

Wroblewski suggested the general form,

 

[x, x+a+b, x+b+d, x+2a+2b, p]k = [x+b, x+d, x+a+2b, x+a+b+d, q]k

 

where, for this case, {a,b,d,p,q,x} = {121, 142, 697, 471, 515, -508}.  This system can be resolved into a final eqn that is only a quartic, though the expressions are messy.  The first author translated the general form into a more symmetric version and found the complete soln in terms of four quadratics to be made squares.  Let,

 

[a+b, -a+b, a+b+2c, a+b-2c, e]k = [b-c-d, b+c-d, b-c+d, b+c+d, f]k     (eq.1)

 

for k = 2,4,6,8, then with {a,b}as free variables, {c,d,e,f} are given as,

 

32c2 = 5a2-10ab+8b2

32d2 = 29a2+6ab+72b2

4e2 = 9a2-6ab+36b2

4f2 = 13a2+2ab+4b2

 

with some {a,b} as trivial like the ratio a/b = {2, 6, etc).  A non-trivial soln is {a,b} = {-288, 43} which makes all {c,d,e,f} as squares and yields the multi-grade which inspired this form.  It is easy to make {c,d} as squares using an elliptic curve, though apparently it is only by chance that {e,f} are squares as well.  But this can provide a rational family to the system,

 

x1k + x2k + x3k + x4k + x5k/2 = y1k + y2k + y3k + y4k + y5k/2for k = 2,4,6,8

 

(End update.)

  
 

8.4 Twelve terms

 

(Update, 1/17/10):  I. x1k+x2k + … + x6k = x1k+x2k + … + x6k , for k = 2,4,8

 

Wroblewski completely solved the system,

 

[ap+d, ap-d, bq+c, bq-c, ar, bs]k = [bp+c, bp-c, aq+d, aq-d, as, br]k,  for k = 2,4,8

 

with the palindromic conditions ma2+nb2 = c2 and na2+mb2 = d2 for some constants {p,q,r,s; m,n} as an elliptic curve imbedded in another elliptic curve.  The soln is,

 

{p,q,r,s; m,n} = {u+v, u-v, u-2v, u+2v; w, v2}

 

where {u,v,w} satisfy the elliptic curve,

 

3u4+20u2v2+16v4 = w2     (eq.1)

 

The first point is {u,v,w} = {2,1,12} which yields {p,q,r,s; m,n} = {3,1,0,4 ; 12, 1} and solving the simultaneous quadratics 12a2+b2 = c2 and a2+12b2 = d2 requires another elliptic curve.  Since r = 0, this in fact is the known (8.5.5) identity discussed in Ten Terms, hence is valid for k = 2,4,6,8.  The second point is {u,v,w} = {6,1,68} which gives {7,5,4,8;  68, 1} so,

 

[7a+d, 7a-d, 5b+c, 5b-c, 4a, 8b]k = [7b+c, 7b-c, 5a+d, 5a-d, 8a, 4b]k

 

for k = 2,4,8 where 68a2+b2 = c2 and a2+68b2 = d2.  And so on for all rational points of eq.1.  (End update.)

 

 

(Update, 1/12/10):  II. x1k+x2k + … + x6k = x1k+x2k + … + x6k , for k = 2,4,6,8

 

Wroblewski gave three new beautiful identities:

 

1.  If 64a2-11b2 = 5c2 and -11a2+64b2 = 5d2, then

 

(2a+5b+c)k + (2a+5b-c)k + (5a-2b+d)k + (5a-2b-d)k + (4a+6b)k + (6a-4b)k =

(2a-5b+c)k + (2a-5b-c)k + (5a+2b+d)k + (5a+2b-d)k + (4a-6b)k + (6a+4b)k

 

for k = 2,4,6,8.  Trivial ratios are a/b = {1, 2}.  These two quadratic conditions define an elliptic curve, hence the system has an infinite number of rational solns.

 

2.  If 248a2-27b2 = 5c2 and -27a2+248b2 = 5d2, then

 

(a+10b+c)k + (a+10b-c)k + (10a-b+d)k + (10a-b-d)k + (a+11b)k + (11a-b)k =

(a-10b+c)k + (a-10b-c)k + (10a+b+d)k + (10a+b-d)k + (a-11b)k + (11a+b)k

 

for k = 2,4,6,8.  Trivial ratios are a/b = {1, 3}, while a non-trivial soln is {a,b} = {9533, 3439} from which an infinite more can be computed.  These belong to the general form, call this Form 1,

 

[pa+qb+c, pa+qb-c, qa-pb+d, qa-pb-d, ra+sb, sa-rb]k =

[pa-qb+c, pa-qb-c, qa+pb+d, qa+pb-d, ra-sb, sa+rb]k

 

where ma2+nb2 = c2 and na2+mb2 = d2 (two quadratics with palindromic coefficients) which also appeared in Tenth Powers by this author.  Notice how the variable b is just negated in the RHS.  Thus, there are now three known non-trivial solns,

 

{p,q,r,s} = {1, 3, 2, 8};  {m,n}= {45, -11};  for k = 2,4,6,8,10

{p,q,r,s} = {2, 5, 4, 6};   {m,n}= {64/5, -11/5}

{p,q,r,s} = {1, 10, 1, 11}; {m,n}= {248/5, -27/5}

 

and whether there are more is unknown, though Wroblewski has done a numerical search for {p,q,r,s} within a bound.  (By expanding the system, one can linearly express {m,n} in terms of {p,q,r,s}, so the latter are the true unknowns.)

 

3.  If -15a2+96b2 = 9c2 and -96a2+825b2 = 9d2, then

 

(a+8b+c)k + (a+8b-c)k + (3a-2b+c)k + (3a-2b-c)+ (2a-b+d)k + (2a-b-d)k  =

(a-8b+c)k + (a-8b-c)k + (3a+2b+c)k + (3a+2b-c)+ (2a+b+d)k + (2a+b-d)k 

 

for k = 2,4,6,8.  Trivial ratios are a/b = {1, 2, 3, 5/2}.

 

This can be generalized into a Form 2,

 

[pa+qb+c, pa+qb-c, ra+sb+c, ra+sb-c, ta+ub+d, ta+ub-d]k =

[pa-qb+c, pa-qb-c,  ra-sb+c,  ra-sb-c,  ta-ub+d,  ta-ub-d]k

 

where p1a2+p2b2 = c2, and p3a2+p4b2 = d2.  Again, the variable b has simply been negated in the RHS.  Notice the terms obey,


x1-x2 = x3-x4 = y1-y2 = y3-y4 


and can be used to form a (k.8.8) for k = 1,3,5,7,9. Whether there are other non-trivial solns of Form 2, or if it can be extended up to k = 10 without being a Form 1, is unknown.

  

 

8.5 Fourteen terms

 

Birck-Sinha Theorem

 

Theorem:  If ak+bk+ck = dk+ek+fk,  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 + 0 =

0 + (2a)k + (2b)k + (2c)k  + (-d+e+f)k +  (d-e+f)k + (d+e-f)k + (d+e+f)k

 

for k = 1,2,4,6,8.  

 

Note that for any {a,b,c}, then,

 

0 + (2a)k + (2b)k + (2c)k = (a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k,  for k = 1,2

 

and likewise for the {d,e,f}.  The zero term is added to show that, if expressed in terms of {xi, yi}, then sums of appropriate terms from opposite sides are equal, namely,

 

x1+y1 = x2+y2 = x3+y3 = x4+y4 = a+b+c

x5+y5 = x6+y6 = x7+y7 = x8+y8 = d+e+f

 

which also indicates that it is just a special case of a more general theorem given in the next section.

 

Proof: (Sinha) Define {s2, s4}:= {a2+b2+c2,  a4+b4+c4}, and Fk:= (a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k - (2a)k - (2b)k - (2c)k 

 

and we find that:  F1 = 0,    F2 = 0,   F4 = 12(s22-2s4),   F6 = 60s2(s22-2s4),    F8 = 28(7s22+2s4)(s22-2s4)

 

Given analogous functions t1, t2 in the variables {d,e,f}, one can evaluate Fk(d,e,f) similarly.  Since it is given that ak+bk+ck = dk+ek+fk for  k = 2,4, then Fk(a,b,c) = Fk(d,e,f) for k = 1,2,4,6,8.  (End proof.)  Also, using F4 and F6, we get a Ramanujan-type identity,

 

5[a2+b2+c2][(a+b+c)4 + (-a+b+c)4 + (a-b+c)4 + (a+b-c)4 - (2a)4 - (2b)4 - (2c)4] =

[(a+b+c)6 + (-a+b+c)6 + (a-b+c)6 + (a+b-c)6 - (2a)6 - (2b)6 - (2c)6]

 

for arbitrary {a,b,c}.  Using a similar analysis, Hirschhorn showed Ramanujan may have used this method to find his 6-10-8 Identity and, in the process, Hirschhorn found an analogous 3-7-10 Identity.  We can also use the Birck-Sinha theorem to find identities with fewer terms.  To recall, we have,

 

(a+b+c)k + (-a+b+c)k + (a-b+c)k + (a+b-c)k + (2d)k + (2e)k + (2f)k + 0 =

0 + (2a)k + (2b)k + (2c)k  + (-d+e+f)k +  (d-e+f)k + (d+e-f)k + (d+e+f)k

 

If we equate a pair of terms (in color) on one side to the other side,

 

(a+b-c) = (2c)

(2f) = (d+e-f)

 

we get, in addition to the requirement ak+bk+ck = dk+ek+fk, for k = {2,4}, the auxiliary conditions that,

 

a+b = 3c

d+e = 3f

 

Fortunately, such a system is solvable by,

 

{a,b,c} = {3p+u, 3p-u, 2p}

{d,e,f} = {3q+v, 3q-v, 2q}

 

(3p+u)k + (3p-u)k + (2p)k = (3q+v)k + (3q-v)k + (2q)k,  k = {2,4}

 

where,

 

p2+12q2 = u2

12p2+q2 = v2

 

and the Birck-Sinha theorem becomes the known (k.5.5) Birck-Sinha Identity discussed previously,

 

(4p)k + (p-u)k + (p+u)k + (2p)k + (3q+v)k + (3q-v)k + (2q)k + 0 =

0 + (3p+u)k + (3p-u)k + (2p)k  + (q-v)k +  (q+v)k + (2q)k + (4q)k

 

for k = {1,2,4,6,8} since two terms (as well as the zero term) on either side cancel out.  One cannot use the theorem to find another (k.5.5) identity (as trying to cancel other combinations of terms yield trivial results). Wroblewski instead looked for more parametric solutions to,

 

x1k + x2k + x3k = y1k + y2k + y3k,  for k = 2,4

 

where at least x1+x2 = 3x3 since this is enough to yield a (k.6.6) identity.  Equivalently, let,

 

{x1, x2, x3} = {3p+u, 3p-u, 2p}

 

and the condition is,

 

y12 + y22 + y32 = 2(11p2+u2)

y14 + y24 + y34 = 2(89p4+54p2u2+u4)

 

though any relation that yields xi = yi must be avoided as it is trivial.  There are only seven infinite families known so far.  The first is by Birck & Sinha, the second is by Piezas, while the last five are by Wroblewski:

 

1. y1+y2 = 3y3:

 

(3p+u)k + (3p-u)k + (2p)k = (3q+v)k + (3q-v)k + (2q)k

 

p2+12q2 = u2

12p2+q2 = v2

 

Ex. {p,q} = {218, 11869}

 

2. x1+x2 = y1+y2:

 

(3p+u)k + (3p-u)k + (2p)k = (3p+v)k + (3p-v)k + (2q)k

 

26p2-3q2 = -u2 

24p2-q2 = -v2

 

Ex. {p,q} = {143, 749}

 

3. x1-x2 = y1-y2:

 

(3p+u)k + (3p-u)k + (2p)k = (u+v)k + (-u+v)k + (2q)k

 

8p2-3q2 = -3u2 

11p2-2q2 = v2

 

Ex. {p,q} = {153, 343}

 

Note:  This yields the nice (k.6.6):

 

(4p)k + (p-u)k + (p+u)k + (u+v)k + (-u+v)k + (2q)k =

(3p+u)k + (3p-u)k  + (q-u)k + (q+u)k + (-q+v)k + (q+v)k,  for k = {1,2,4,6,8}

 

4. 3(x1-x2) = 2(y1-y2):

 

(3p+2u)k + (3p-2u)k + (2p)k = (3u+v)k + (-3u+v)k + (2q)k

 

8p2-3q2 = 5u2 

3p2+q2 = v2

 

Ex. {p,q} = {323, 397}

 

OR,

 

p2-q2 = -9u2 

104p2-23q2 = 9v2

 

Ex. {p,q} = {208, 233}

 

5. x1-x2 = 3y1-2y2:

 

(3p+u)k + (3p-u)k + (2p)k = (2u-2v)k + (2u-3v)k + (2q)k

 

33u2-134uv+133v2 = 153p2

-192u2+112uv+937v2 = 612q2

 

Ex. {u,v} = {7,3}

 

6. 2x1-x2 = 3(y1-y2):

 

(3p+u)k + (3p-u)k + (2p)k = (1/2)k ((p+u+v)k + (-p-u+v)k + (2q)k)

 

19p2+22pu+3u2 = 3q2 

91p2-50pu+3u2 = 3v2 

 

Ex. {p,u} = {9, -8}, or  {p,u} = {45, 76}

 

7. 4x1+x2 = 3(y1-y2):

 

(3p+u)k + (3p-u)k + (2p)k = (1/2)k ((5p+u+v)k + (-5p-u+v)k + (2q)k)

 

67p2+38pu+3u2 = 3q2 

77p2+106pu-3u2 = -3v2 

 

Ex. {p,u} = {9, -8}

 

The two quadratic conditions of each family define an elliptic curve and, from an initial rational point, one can find an infinite more.

 

 Q. Any others?  For the latest data on higher powers with a restricted number of terms, see Wroblewski's "A Collection of Numerical Solutions of Multi-grade Equations".  (End update.)

 

 

 

(Update, 1/17/10):  Wroblewski

 

The Birck-Sinha Theorem can be expressed in the form,

 

(p+q)k + (-p+q)k + (-q+v)k + (q+v)k + (-r+u)k + (r+u)k + (2s)k + 0 =

(-p+v)k + (p+v)k + (2q)k + 0 + (r+s)k + (-r+s)k + (-s+u)k + (s+u)k

 

This still has the expected common sums,

 

x1+y1 = x2+y2 = x3+y3 = x4+y4 = q+v

x5+y5 = x6+y6 = x7+y7 = x8+y8 = s+u

 

By letting,

 

{p,q,r,s} = {-a+b, c, -d+e, f};   {u,v} = {d+e, a+b}

 

one recovers the Birck-Sinha Theorem.  Wroblewski gave the special case such that the above is valid for k = 1,2,4,6,8 as,

 

ps = qr

p2+3q2+s2 = u2

q2+r2+3s2 = v2

 

though there are certain {p,q,r,s} such that the system is trivial.  Proof:  Simply solve for {u,v}, express s in terms of {p,q,r}, and substitute into the system for k = 1,2,4,6,8.  Extending to k = 10 will give the trivial values.  <End proof> 

 

Assuming r = pt, s = qt, it reduces to two conditions,

 

p2+3q2+(qt)2 = u2

q2+(pt)2+3(qt)2 = v2

 

which are two quadratic polynomials to be made squares and can easily be solved as an elliptic curve.  An example is {p,q,t} = {9, 8, 2}.  Alternatively, by assuming p = 3qr = 3s, this equates two terms of one side to the other side, {-p+q, -r+s} = {-2q, -2s}.  The conditions become,

 

12q2+s2 = u2

q2+12s2 = v2

 

Since two terms from each side cancel out, this reduces from a (k.7.7) to the (k.5.5) Birck-Sinha Identity discussed in a previous section.  (End update.)

 

 

(Update, 2/2/10):  Piezas

 

More generally, given the same form above,

 

(p+q)k + (-p+q)k + (-q+v)k + (q+v)k + (-r+u)k + (r+u)k + (2s)k + 0 =

(-p+v)k + (p+v)k + (2q)k + 0 + (r+s)k + (-r+s)k + (-s+u)k + (s+u)k

 

for k = 1,2,4,6,8, then {p,q,r,s,u,v} must satisfy two simultaneous eqns,

 

p2+2q2+v2 = r2+2s2+u2                  (eq.1)

(p2-q2)(q2-v2) = (r2-s2)(s2-u2)         (eq.2)

 

though avoiding trivial cases.  Proof:  Simply solve for {u,v} in terms of {p,q,r,s}, and substitute into the system.  Also, this can be shown as the Birck-Sinha theorem in disguise by letting {p,q,r,s} = {-a+b, c, -d+e, f};  {u,v} = {d+e, a+b}, and (eq.1) and (eq.2) are satisfied if,

 

ak+bk+ck = dk+ek+fk,   for k = 2,4              

 

For the special case u = v, then eq.1 becomes,

 

p2+2q2 = r2+2s2

 

the complete soln of which reduces eq.2 to the easy problem of making a quadratic polynomial into a square.  (End update.)

 

   

8.6  Sixteen terms

 

Piezas

 

Theorem 1. If ak+bk+ck+dk = ek+fk+gk+hk, for k = 2,4 and abcd = efgh, where a+b+c+d ≠ 0 and e+f+g+h ≠ 0, (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 =

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

 

for k = 1,2,4,6,8.  (Call this derived system V2.)

 

This has the common sums,

 

x1+y1 = x2+y2 = x3+y3 = x4+y4 = a+b+c+d

x5+y5 = x6+y6 = x7+y7 = x8+y8 = e+f+g+h

 

Thus if a+b+c+d = e+f+g+h = 0, then xi = -yi and the system becomes trivial so this case should be avoided.  However, one can set d = h = 0, and this (k.8.8) reduces to the (k.7.7) Birck-Sinha Theorem.

 

Proof:  Following Sinha’s approach, define,

 

{s1, s2, s4} = {abcd, a2+b2+c2+d2,  a4+b4+c4+d4} and Fk = (a+b+c-d)k + (a+b-c+d)k + (a-b+c+d)k + (-a+b+c+d)k - (2a)k - (2b)k - (2c)k - (2d)k

 

and we can evaluate Fk for certain k in terms of the si.  Set s0 = -8s1+s22-2s4, then,

 

F1 = F2 = 0,    F4 = 12s0,    F6 = 60s0s2,    F8 = 28s0(-24s1+7s22+2s4),  

 

For the variables e,f,g,h, assume analogous functions t1, t2, t4, t6.  Obviously Fk(e,f,g,h) at the same k will be evaluated identically.  Since it is given that ak+bk+ck+dk = ek+fk+gk+hk, for k = 2,4, and abcd = efgh, this implies Fk(a,b,c,d) = Fk(e,f,g,h) for k = 1,2,4,6,8, and we complete the proof.  Also, using F4 and F6, we can get the more general identity,

 

5[a2+b2+c2+d2]*

[(a+b+c-d)4 + (a+b-c+d)4 + (a-b+c+d)4 + (-a+b+c+d)4 - (2a)4 - (2b)4 - (2c)4 - (2d)4] =

[(a+b+c-d)6 + (a+b-c+d)6 + (a-b+c+d)6 + (-a+b+c+d)6 - (2a)6 - (2b)6 - (2c)6 - (2d)6]

 

for arbitrary {a,b,c,d}.  

 

Note:  The theorem can be extended to 10th powers if,

 

ak+bk+ck+dk = ek+fk+gk+hk, for k = 2,4,6

 

and abcd = efgh, but this has only trivial solns.  Proof:  Define Fk = ak+bk+ck+dk-(ek+fk+gk+hk).  If Fk = 0 for k = 2,4, then,

 

F8 = -4(abcd+efgh)(abcd-efgh) + (4/3)(F6)(a2+b2+c2+d2)

 

Thus, if both abcd-efgh = 0 and F6 = 0, this implies F8 = 0 as well, in contradiction to Bastien’s Theorem.  So, if Fk = 0 for k = 2,4,6, then abcd ≠ efgh. Or if abcd = efgh, then Fk = 0 only for k = 2,4. <End proof.> 

 

  

(Update, 2/15/10):  Piezas, Wroblewski

 

Using a numerical search with constraints, Wroblewski found hundreds of solns to a multi-grade octic chain of form,

 

x1k+x2k+x3k+x4k-(x5k+x6k+x7k+x8k) = y1k+y2k+y3k+y4k-(y5k+y6k+y7k+y8k) = z1k+z2k+z3k+z4k-(z5k+z6k+z7k+z8k)

 

valid for k = 2,4,6,8.  An example is,

 

[221, 93, 73, 9]k-[219, 117, 31, 17]k = [198, 116, 96, 32]k-[192, 144, 58, 44]k = [189, 125, 105, 23]k-[175, 161, 75, 27]k

 

Note that it is also true x1k+x2k+x3k+x4k = x5k+x6k+x7k+x8k, for k = 2,4 and x1x2x3x4 = x5x6x7x8, relationships which also hold for the yi and zi.  It turns out the general case could be explained in the context of the theorem given by the author in section 8.6 above.  Moving half of the terms of one side to the other yields,

 

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

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

 

and since either sign of {±d, ±h} can be used and still satisfy abcd = efgh, then the RHS has two distinct octuplets for a given {a,b,c,d,e,f,g,h}.  Thus, given,

 

p1k+p2k+p3k+p4k = q1k+q2k+q3k+q4k, k  = 2,4

 

where p1p2p3p4 = q1q2q3q4, then this automatically leads to an octic chain of length 3.  Wroblewski’s solns satisfies the constraints,

 

p1p2 = q3q4

p3p4 = q1q2

p12+p22 = q12+q22

p32+p42 = q32+q42

p14+p24+p34+p44 = q14+q24+q34+q44

 

The first author found the complete soln to this system as an elliptic “curve”, namely,

 

{p1, p2, p3, p4} = {ar+bc, ac-br, ds+ef, df-es}

{q1, q2, q3, q4} = {ar-bc, ac+br, ds-ef, df+es}

 

r = (abc2+def2)(d2-e2)(f/y),  s = (abc2+def2)(a2-b2)(c/y)

 

and {a,b,c,d,e,f} must satisfy,

 

(abc2+def2)(c2de)(a2-b2)2 + (abc2+def2)(abf2)(d2-e2)2  = y2

 

which is only a quartic polynomial in {c,f} to be made a square.  Given one soln, then an infinite more can be found.

 

Piezas

 

Another set of conditions is,

 

p1p4 = q1q4

p2p3 = q2q3

p1+p2 = q1+q2

p1k+p2k+p3k+p4k = q1k+q2k+q3k+q4k,  k = 2,4

 

The soln can be given as,

 

{p1, p2, p3, p4} = {a(p+q), b(r-s), c(r+s), d(p-q)}

{q1, q2, q3, q4} = {a(p-q), b(r+s), c(r-s), d(p+q)}

 

{p,q,r,s} = {a(b2-c2),  be,  (a2-d2)b,  ae}

 

and where {a,b,c,d,e} satisfy,

 

a2(3b2-c2)+e2 = (b2+c2)d2

 

This can be easily solved parametrically as quadratic forms as discussed in Form 17, Fourth Powers .  (End update)

  

 

(Update, 1/25/10):  Wroblewski

 

Given,

 

(a2)k + (b2)k + (a2+b2)k = (c2)k + (d2)k + (c2+d2)k,  for k = 2,4    (eq.1)

 

or equivalently,

 

a4+a2b2+b4 = c4+c2d2+d4

 

then we get the [k.8.8] identity,

 

[ap+cq, -ap+cq, bp+dq, -bp+dq, dp+aq, dp-aq, cp+bq, cp-bq]k =  

[ap+dq, -ap+dq, bp+cq, -bp+cq, cp+aq, cp-aq, dp+bq, dp-bq]k,   for k = 1,2,4,8

 

(Notice how {c,d} of the LHS is replaced by {d,c} in the RHS.)  for arbitrary {p,q} and some constants {a,b,c,d}, hence giving an identity in terms of linear forms.  An example is {a,b,c,d} = {1, 26, 17, 22}, though Choudhry has given a 7th deg identity for,

 

a4+ma2b2+b4 = c4+mc2d2+d4

 

for arbitrary m.  See Form 3 here.  To find a [8.7.7], since {p,q} are arbitrary, one can equate appropriate xp = yq so a pair of terms cancel out.  Note:  Eq. 1, or more generally its unsquared version, is at the core of the Ramanujan 6-10-8 Identity.  (End update.)

 

 

(Update, 1/17/10):

8.5 Eighteen terms

 

Lander

 

(28k+4 + 1)8 = (28k+4-1)8 + (27k+4)8 + (2k+1)8 + 7(25k+3)8 + 7(23k+2)8

 

Note:  This has a counterpart for 4th powers.  I misplaced my notes but, if I remember correctly, 7 is replaced by 3(?), and the powers of 2 are changed appropriately.  If someone can come up with something, pls submit it.  (End update)

 

(Update, 1/18/10):  Wroblewski suggested the family which starts with Pythagorean triples,

 

(2xy)2 + (x2-y2)2 = (x2+y2)2

 

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

 

(16x7y)8 + (2xy7)8 + 7(8x5y3)8 + 7(4x3y5)8 + (16x8-y8)8 = (16x8+y8)8

 

where Lander’s 8th powers identity was just  the special case {x,y} = {2k, 1}.  If we continue and try to decompose (256x16+1)16 - (256x16-1)16 as a sum of 16th powers, we find the coefficients have prime factors other than 2 and/or 2k-1, hence such neat decompositions of nth = 2k powers stop at n = 8.   (End update)

 

 

Previous Page        Next Page

Comments