030: Tenth Powers

 
 
PART 13.  Tenth Powers
 
 
10.1  Ten terms
10.2  Twelve Terms
10.3  Fourteen Terms
10.4  Sixteen Terms
10.5  Twenty Terms
 
 
10.1  Ten Terms 
 
No non-trivial solution in the rationals is yet known to equal sums of five 10th powers,
 

x110+x210+x310+x410+x510 = y110+y210+y310+y410+y510

 
 
10.2  Twelve Terms
 
The eqn,
 

x1k+x2k+x3k+x4k+x5k+x6k = y1k+y2k+y3k+y4k+y5k+x6k     (eq.1)

 
for k = 10 has relatively small solns.  The first one was found by Randy Ekl in 1997,
 
[5, 23, 34, 34, 85, 92] = [16, 25, 28, 32, 71, 95]
 
There are many more and whether these belong to a parametric family is unknown.  However, a family has been found for the even multi-grade system k = 2,4,6,8,10 (pls see update below).  A numerical example (not belonging to the known family) was first found in 1999,

 

{22, 61, 86, 127, 140, 151} = {35, 47, 94, 121, 148, 146}

 

by Kuosa and Meyrignac as k = 10, and Shuwen (who noticed it was not just for k = 10 but was multi-grade) which automatically leads to an ideal soln of deg 11.  To see if this is also for k = 1 if the terms are signed, there are now 64 possible sums of ± a ± b ± c ± d ± e ± f  (and an equal number for the terms on the other side), and Shuwen found that sums of one side are equal to the other in seven ways, three of which are symmetric,


a-b-c+d-e+f = g-h-i+j-m+n = 13

a+b+c-d-e+f = g+h+i-j-m+n = 53

a+b-c-d+e+f = g+h-i-j+m+n = 161

 

a+b+c-d+e+f = -g+h-i+j+m+n = 333

a+b-c+d-e+f = -g-h+i+j-m+n = 135

a-b+c+d-e+f = -g-h+i-j+m+n = 185

a-b+c-d+e-f = g-h+i+j-m-n = -91


not counting seven other combinations which simply negate the terms.  Also, similar to the known family, it is also the case that,


x32+x4 = y22+y62


Whether this is: (1) a fluke, or (2) a property of the possible family which this example belongs to, or (3) solns to even systems with signed terms tend to be valid for k = 1 the more terms there are, is not known.  While there are parametric (k.6.6) with all three side-conditions,

 

x1 ± x2 = y1 ± y2

x3 ± x4 = y3 ± y4

x5 ± x6 = y5 ± y6

 

for sextic and octic systems, it is not known if it can be for the decic. (The Choudhry-Wroblewski decic family uses only two.)  


Note:  Kuosa and Meyrignac were looking at 10.6.6 and among the 14 solns they found, one turned out to be for k = 2,4,6,8,10.  If this was just a statistical fluke, one might expect some of the others to be also valid for, say, k = 2 or 4.  This author checked and none is good for any other even exponent.  So there is the situation of a small soln that is suddenly multi-grade for six exponents, counting k = 1.  Together with its other properties, it seems there might be more to it than meets the eye.

 
Update (June 6, 2009):  Choudhry, Wroblewski
 
It has been brought to my attention (by J. Wroblewski) that the case of an equal sum of six tenth powers,

 

a1k+a2k+a3k+a4k+a5k+a6k = b1k+b2k+b3k+b4k+b5k+b6k,  for k = 2,4,6,8,10

 

has already been cracked (in 2008) and shown to have an infinity of solns by reducing it, like the other systems, to a particular elliptic curve, though the Kuusa-Meyrignac-Shuwen soln does not belong to this family.  (See their paper "Ideal Solutions of the Tarry-Escott Problem of Degree Eleven with Applications to Sums of Thirteenth Powers", Hardy-Ramanujan Journal, Vol. 31, 2008.)  Choudhry and Wroblewski defined ai and bi as, 

 

a1 = 2xy+x+2y-7;   b1 = 2xy+2x+y-7

a2 = 2xy-x-2y-7;     b2 = 2xy-2x-y-7

a3 = 2xy-2x+y+7;    b3 = 2xy-x+2y+7

a4 = 2xy+2x-y+7;    b4 = 2xy+x-2y+7

a5 = 3x+5y;             b5 = 5x+3y

a6 = 5x-3y;              b6 = 3x-5y

 

where a variable z has been set z = 1 without loss of generality.  Substituting these values into the system, the higher eqns hold if the ff condition is satisfied,

 

8x2y2-17x2-17y2+98 = 0

 

Do the simple change of variables {x, y} = {u, v/(8u2-17)} and the condition reduces to,

 

(-17+8u2)(-98+17u2) = v2

 

or a quartic polynomial in u that is to be made a square.  As Choudhry and Wroblewski proved, this is an elliptic curve.  One trivial soln is u = 1 but from this initial point, we can compute an infinite number of non-trivial ones such as u = -457/353, etc.  However, by looking at the solns ai and bi, one can see certain relationships and symmetries between them.  Using those relationships, this author will show that it can be simplified to a form analogous to the Letac-Sinha identity for k = 2,4,6,8.
 
 
(Update, 2/1/10):  Choudhry, Wroblewski  

 

The 6th power (k.4.4) identity can be generalized to the form (k.6.6) by adding the italized pair of terms,

 

[-(axy+bx+cy-d), axy-bx-cy-d, axy-cx+by+d, -(axy+cx-by+d), ex+fy, fx-ey]k =

[-(axy+by+cx-d), axy-by-cx-d, axy-cy+bx+d, -(axy+cy-bx+d), ey+fx, ex-fy]k

 

where again, {x,y} merely swap places.  (Since the eqn is homogeneous, one may assume the leading coefficient of the terms as a = 1 without loss of generality.)  This obeys the same basic constraints (up to sign changes),

 

p1-p2 = q1-q2,

p3-p4 = q3-q4,

(-p1)k+p2k+p3k+(-p4)k = (-q1)k +q2k+q3k+(-q4)k,  k = 1,2

p52+p62 = q52+q62

 

If we assume,

 

p1+p2+p5 = q1+q2+q5, 

p3+p4+p6 = q3+q4+q6, 

 

then it must be the case that 2(b-c) = e-f   (eq.1).  Choudhry and Wroblewski found for k = 2,4,6,8,10, the equivalent non-trivial solns,

 

{a,b,c,d,e,f} = {1, 1, 2, 14, 3, 5},  where 2x2y2-17(x2+y2)+392 = 0

{a,b,c,d,e,f} = {2, 1, 2, 7, 3, 5},  where 8x2y2-17(x2+y2)+98 = 0

 

Disregarding eq.1, two other distinct non-trivial solns found (by Piezas), though only up to k = 2,4,6,8, are,

 

{a,b,c,d,e,f} = {5, 3, 7, 3, 2, 10},  where 125x2y2-53(x2+y2)+45 = 0

{a,b,c,d,e,f} = {5, 9, 11, 11, 10, 12}, where 125x2y2-221(x2+y2)+605 = 0

 

though this author is not certain if these are the same 8th power multi-grades found by Wroblewski.  See also Sixth Powers for the context of this identity.  (End update.)

 
 
Piezas

 

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

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

 

for k = 2,4,6,8,10,  where 45a2-11b2 = c2 and -11a2+45b2 = d2. 

 

where the ratio a/b = {2, 1/2, 3/2, 2/3} must be avoided as it yields trivial solns.  Labelling terms as ai and bi, then a small tweak in the sign of {a1, a4, b1, b4} can make it valid for k = 1 as well.  Note that, just like in the Letac-Sinha identity, the two quadratic polynomials to be made squares have palindromic or reversible coefficients. One can easily verify this soln by solving {c,d} in radicals for arbitrary {a,b} and substituting them into the system.  Of course, if they are to be rational, then appropriate {a,b} must be chosen, with the smallest non-trivial one being {a,b} = {186, 331} giving,
 
[886, -293, 1180, 953, 1510, -413] = [700, -107, 1511, 622, 1138, -1075] 
 
after removing a small common factor.  By solving the first condition as a quadratic form {a, b} = {u2+11v2, 2u2+2uv-22v2} using this on the other will result in a quartic polynomial that is to be made a square, with some {u,v} as trivial, but a non-trivial is {u,v} = {12, -5}. Treating this as an elliptic curve, from this initial point, an infinite number of rational solns can then be computed.  This identity was found using the relationships between the ai and bi and, whether in the Choudhry-Wroblewski version or by this author, they satisfy seven side conditions: 
 
a1+a2 = b1+b2

a3+a4 = b3+b4

a1-a2-a5 = b1-b2-b5

a3-a4+a6 = b3-b4+b6

a1+a2+a3-a4+a5+a6 = b1+b2-b3+b4+b5-b6

a1k+a2k+a3k+a4k = b1k+b2k+b3k+b4k,  for k = 1,2

 

and together with the five eqns,

 

a1k+a2k+a3k+a4k+a5k+a6k = b1k+b2k+b3k+b4k+b5k+b6k,  for k = 2,4,6,8,10

 

we have 12 (or 11) eqns in 12 unknowns, call this augmented system as S10+.  By solving this system, it can be shown that the identity given by this author is the only non-trivial soln to S10+.  (There may be just 11 eqns as one may be a consequence of the others.) 
 
Note 1:  By negating {a1, a4, b1, b4}, the identity in fact is also valid for k = 1.  Also, by eliminating or changing one of the side conditions, it may be possible to come up with a different identity from this one.  What would be an appropriate change, I do not yet know.
Note 2:  I guess my postscript at the final section need to be modified. (However, the case k = 1,3,5,7,9 still needs to be cracked.)
 
(Update, 1/12/10):  The identity has the form,
 
[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.  Notice how the variable b is just negated in the RHS.  Wroblewski did a search and found two new solns, though only for k = 2,4,6,8:

 

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

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

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

 

See also Eighth Powers.  (End update.)

 
 
(Update, 1/11/10)
 
10.3  Fourteen Terms
 

The smallest soln to,

 

x1k+x2k+x3k+x4k+x5k+x6k+x7k = y1k+y2k+y3k+y4k+y5k+y6k+y7k    (eq.1)

 

for k = 10 is,

 

[1, 8, 15, 26, 26, 33, 38]10 = [22, 23, 24, 29, 32, 35, 36]10

 

given by Ekl (“New results in equal sums of like powers”, 1998).  However, as far back in 1939, Moessner already managed to solve eq.1 (pre-computer searching!).  The second(?) smallest soln, and the smallest in distinct integers, turns out to be multigrade,

 

[32, -1, -61, -55, -31, -13, -28, 68]k

[44, -13, -49, -67, 20, -64, 23, 17]k

 

for k = 1,2,4,6,8,10, and belongs to an infinite family.  Note that its terms obey the relations,

 

-32+44 = -1+13 = 61-49 = -55+67 = 12

31+20 = -13+64 = 28+23 = 68-17 = 51

 

Wroblewski, Piezas

 

Let a2+b2 = c2 and a2+52b2 = d2, a pair of conditions which is a concordant form. Then a (k,7,7) for k = 1,2,4,6,8,10 is,

 

(8b)k + (5a-4b)k + (-a-2d)k +  (a-2d)k + (-5a-4b)k + (a-4b)k + (-12b+4c)k + (12b+4c)k =

(4a+8b)k + (a-4b)k + (3a-2d)k + (-3a-2d)k + (-4a+8b)k + (-16b)k + (a+4c)k + (-a+4c)k

 

With terms as {xi, yi}, the identity obeys the constraints,

 

x1-y1 = -(x2-y2) = x3-y3 = -(x4-y4) = -4a

x5-y5 = -(x6-y6) = x7-y7 = -(x8-y8) = -(a+12b)

 

This was derived from a (k.8.8) discussed in the next section but since two terms (in blue) cancel out, then it reduces to a (k.7.7).  However, ratios such as a/b = {4, 12, 12/5, 4/3} should be avoided as the system becomes trivial, with the smallest non-trivial soln as the smallest Pythagorean triple {a,b,c} = {3, 4, 5}, and d = 29.  This yields,
 
[32, -1, -61, -55, -31, -13, -28, 68]k = [44, -13, -49, -67, 20, -64, 23, 17]k  
 
This was cited by Ekl (1998) as the smallest soln in distinct integers, but it was not mentioned in the paper that it was also multigrade.  It is quite interesting that the smallest multigrade (k.7.7) for k = 2,4,6,8,10 belongs to a family, as it is not known if the smallest multigrade (k.6.6) belongs to a family or not.

 

The two quadratic polynomials generally define an elliptic curve, and there is an infinite number of solns, with other small {a,b} as {612, 35} and {783, 56}.  A pair of algebraic forms {a2+b2, a2+Nb2} that is to made squares is called a concordant form.   (However, found throughout this book, is the more general case of  {pa2+qb2, ra2+sb2} to be made squares.)  This identity is a special case of a more general one involving 16 terms given below.

 

 
10.4  Sixteen Terms
 

Piezas, Wroblewski

 

The latter author suggested the system S1,

 

[-a-c+x, a+c+x, -a+b+c-y, a-b-c-y, -a-b-c-t, a+b+c-t, -b+c+z, b-c+z]k =

[a-c+x, -a+c+x,  a+b+c-y, -a-b-c-y, -a+b-c-t, a-b+c-t, b+c+z, -b-c+z]k           

 

which, for some {a,b,c,x,y,z,t}, is valid for k = 1,2,4,6,8,10.  This has a lot of symmetries and its terms {xi, yi} obeys the constraints,

 

x1-y1 = -(x2-y2) = x3-y3 = -(x4-y4) = -2a

x5-y5 = -(x6-y6) = x7-y7 = -(x8-y8) = -2b

x1-x2+y1-y2 = -x3+x4+y5-y6 = -x7+x8-y7+y8 = -4c

x5-x6 = -y3+y4 = -2(a+b+c)

 

An example by Wroblewski is,

 

[-31, 193, 75, -100, -179, 164, 72, 51]k = [53, 109, 159, -184, -60, 45, 191, -68]k

 

which is for k = 1,2,3,4,6,8,10.  The former author found the complete radical soln using Mathematica.  This entails making four quartic polynomials as squares.  Set a = 1 without loss of generality and define {x,y,z,t} = {√p, √q, √r, √s}.  Then {p,q,r,s} can be rationally expressed in terms of the free variables {b,c}.  As the actual expressions are unwieldy, first define,

 

p-s = (-2b+9b2+8c+2bc-8c2)/3

q-s = 8(1-b)(1-c)(-b+c)c/(3b+3c+3bc)

q-r = (-9+2b-2c-8bc+8c2)/3

 

and it is easy to solve for {p,q,r}.  Substituting these into the system and factoring at either k = 6,8,10, one will find a common linear factor for the last unknown s.  Expanding for k = 12 will yield the trivial forms of {b,c}.  A non-trivial one which makes {p,q,r,s} squares is {b,c} = {17/12, 5/3} which gives the numerical example above.  However, Wroblewski found that if b+c = 1/2, such as,

 

{a,b,c} = {1, (u+12v)/(4u), (u-12v)/(4u)} 

 

then {p,s} are squares, {q,r} become just quadratic polynomials, and two terms {x6, y2} are identical, so it reduces to the simpler one in section 10.3. (End update)

 
 
(Update, 1/17/10)
 
10.5  Twenty Terms
 
Wroblewski added four more terms to each side of the (k.6.6) to get the (k.10.10),
 
[pa+qb+c, pa+qb-c, qa-pb+d, qa-pb-d, ra+sb, sa-rb, ta+ub, ua-tb, va+wb, wa-vb]k =

[pa-qb+c, pa-qb-c, qa+pb+d, qa+pb-d, ra-sb, sa+rb, ta-ub, ua+tb, va-wb, wa+vb]k

 

for k = 2,4,6,8,10, where ma2+nb2 = c2 and na2+mb2 = d2.  Again, the variable b is just negated in the RHS.  Wroblewski found three non-trivial solns,

 

{p,q,r,s,t,u,v,w; m,n} = {4, 4, 3, 11, 5, 7, 7, -3; 45, -11}

{p,q,r,s,t,u,v,w; m,n} = {4, 4, 9, 17, 15, 13, 13, 7; 189, 85}

{p,q,r,s,t,u,v,w; m,n} = {3, 3, 11, 17, 16, 14, 14, 10; 220, 136}

 

though certain ratios a/b must be avoided and which can determined by expanding for k = 12.  Interestingly, the first (k.10.10) soln uses the same {m,n} as the one for (k.6.6)!  As these two quadratic conditions define an elliptic curve, it is unknown why certain curves are "favored", appearing again and again (it also  appears in a (k.4.4) for k = 1,2,4,6).  It is also remains to be known whether: a) there are others, b) if there is w = v = 0 so reduces to 16 terms, or c) if it can be non-trivially valid for k = 12.  (End update)

 

(Update, 3/1/10):  Piezas

 

This author noticed that, other than the obvious relations p = q, u = v, the above also obey,

 

r = -q+v

s = q+v

mn = rstw

 

Combined with the conditions k = 2,4,6,8,10, these were enough to find a formula for these variables.  Thus,

 

{p,q,r,s; t,u,v,w} = {y, y, -y+z, y+z; x+y, z, z, x-y}

{m,n} = {x2+yz+y2,  x2-yz+y2}

 

where {x,y,z} satisfies the simple quadratic condition x2+3y2 = z2.  Wroblewski’s were the cases,

 

{x,y} = {1, 4}; {11, 4}; {13, 3}

 

However, there is an infinite number of them given by {x,y,z} = {e2-3f2, 2ef, e2+3f2}.  So a fourth one using {x,y} = {11, 5} gives,

 

{p,q,r,s; t,u,v,w; m,n} = {5, 5, 9, 19;  16, 14, 14, 6;  216, 76}

 

with one soln to 216a2+76b2 = c2 and 76a2+216b2 = d2 as,

 

{a,b} = {495753715, 352750681}

 

though presumably smaller solns may exist.  From this initial rational point, one can then compute an infinite more.  And so on for an infinite number of other formulas using different {m,n} and other elliptic curves, though expanding this family at k = 12 or 14 generates only trivial solns.  (End update.)

  

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