001c: Assorted Identities, Part 3 (Equal Sums of Like Powers)

 
 

  

 
(Update, 8/7/09):  V. Equal Sums of Like Powers

 

The results in this section are by the author.  See also the separate section on Equal Sums of Like Powers and those on higher powers.

 

I.  Even Powers (Part 1)

 

Theorem 1:  Define Ek:= ak+bk+ck-(dk+ek+fk).  Let {u,v} = {a2+b2+c2, a4+b4+c4}.  If Ek = 0 for k = 2,4 then,

 

a6+b6+c6-d6-e6-f6 = ± 3(a2-d2)(b2-d2)(c2-d2)

3(a8+b8+c8-d8-e8-f8) = 4(a6+b6+c6-d6-e6-f6) (a2+b2+c2)

6(a10+b10+c10-d10-e10-f10) = 5(a6+b6+c6-d6-e6-f6) ((a2+b2+c2)2+a4+b4+c4)

 

Note:  The d in the RHS of the first eqn can be either {d,e,f}.  And the ± sign depends on what side of the eq is chosen to be {a,b,c} or {d,e,f}. Small soln:

[6, 23, 25]k = [10, 19, 27]k,  for k = 2,4. 

 

Corollary:  These relations can be woven together.  Define n as, a4+b4+c4 = n(a2+b2+c2)2 then,

 

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

 

Proof:  We can concisely express the last two of the eqns above as,

 

3 E8  = 4 E6(u)              (eq.1)

6 E10 = 5 E6(u2+v)        (eq.2)

 

Square eq.1, multiply eq.2 by E6, and then eliminate E62 between eq.1 and eq.2 to get,

 

32 E6E10 = 15 E82 (1+v/u2),  or,

32 E6E10 = 15 E82 (1+n)

 

as stated.  (End of proof).  For the special case when a+b = ± c, since it is true that,

 

a4+b4+(a+b)4 = (1/2)(a2+b2+(a+b)2)2

 

then n = 1/2 and the above becomes,

 

64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10)  = 45(a8+b8+c8-d8-e8-f8)2

 

which is a concise version of the Ramanujan 6-10-8 Identity.  The general case is in fact the first in an infinite family to be given in the following theorems.

 

Theorem 2:  Define Fk:= ak+bk+ck+dk-(ek+fk+gk+hk), and qk:= ak+bk+ck+dkLet {u,v,w} = {q2, q4, q6}.  If Fk = 0 for k = 2,4,6 then,

 

F8 = ± 4(a2-e2)(b2-e2)(c2-e2)(d2-e2)

4 F10 = 5 F8 (u)

8 F12 = 6 F8 (u2+v)

24 F14 = 7 F8 (u3+3uv+2w)

 

Corollary:  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

 

or, equivalently, 25 F8F12 = 12(n+1)(F10)2.  Small soln:  [2, 16, 21, 25]k = [5, 14, 23, 24]k,  k = 2,4,6.

 

Theorem 3:  Define Gk:= ak+bk+ck+dk+ek-(fk+gk+hk+ik+jk) and rk:= ak+bk+ck+dk+ek.  Let {u,v,w,x} = {r2, r4, r6, r8}.  If Gk = 0 for k = 2,4,6,8 then,

 

G10 = ± 5(a2-f2)(b2-f2)(c2-f2)(d2-f2)(e2-f2)

5 G12 = 6 G10 (u)

10 G14 = 7 G10 (u2+v)

30 G16 = 8 G10 (u3+3uv+2w)

120 G18 = 9 G10 (u4+6u2v+3v2+8uw+6x)

 

Corollary:  "Define n as a4+b4+c4+d4+e4 = n(a2+b2+c2+d2+e2)2, then 72 G10G14 = 35(n+1)(G12)2.Small soln:  [71, 131, 180, 307, 308]k = [99, 100, 188, 301, 313]k,  k = 2,4,6,8.

 

Theorem 4:  Define Hk:= ak+bk+ck+dk+ek+fk-(gk+hk+ik+jk+lk+mk) and sk:= ak+bk+ck+dk+ek+fk.  Let {u,v,w,x,y} = {s2, s4, s6, s8, s10}.  If Hk = 0 for k = 2,4,6,8,10, then,

 

H12 = ± 6(a2-g2)(b2-g2)(c2-g2)(d2-g2)(e2-g2)(f2-g2)

6 H14 = 7 H12 (u)

12 H16 = 8 H12 (u2+v)

36 H18 = 9 H12 (u3+3uv+2w)

144 H20 = 10 H12 (u4+6u2v+3v2+8uw+6x)

720 H22 = 11 H12 (Poly1),  where Poly1 is a 5th deg polynomial in terms of u,v,w,x,y. 

 

Corollary:  "Define n as a4+b4+c4+d4+e4+f4 = n(a2+b2+c2+d2+e2+f2)2,  then 49 H12H16 = 24(n+1)(H14)2."  Small soln: [22, 61, 86, 127, 140, 151]k = [35, 47, 94, 121, 146, 148]k,  k = 2,4,6,8,10. 

 

And so on for other k = 2,4,..2n. While an infinite number of solns are known for Theorems 1 to 4, none are known for higher systems.  (Q: Anyone with the programming skills and computing power to find a few?)

 

Note 1:  The coefficient of the Hi in Theorem 4 as {6, 12, 36, 144, 720} divided by 6 is equal to {1, 2, 6, 24, 120}, or the sequence of factorial n!, so it is easy to extrapolate that the next system involves the sequence 7(n!).

 

Note 2:  As an overview, we have the sequence of corollary relations,

 

32 E6E10 = 15(n+1)(E8)2

50 F8F12 = 24(n+1)(F10)2

72 G10G14 = 35(n+1)(G12)2

98 H12H16 = 48(n+1)(H14)2

 

This has the general form,

 

2m2 X2(m-1) X2(m+1) = (m2-1)(n+1)(X2m)2

 

starting with m = 4,5,6… so one can see the next in the series.  However, by taking square roots of the variables, these theorems are also valid for systems with k = 1,2,3,…n, or those satisfying the Prouhet-Tarry-Escott problem.  See also the section on Equal Sums of Like Powers.

 

 

II. Even Powers (Part 2)

 

Theorem 1:  "Define, 

 

{a,b,c,d} = {m+p, m-p, m+q, m-q}  

 

If p2+q2 = m2, then 9(a2+b2-c2-d2)(a4+b4-c4-d4) = 7(a3+b3-c3-d3)2."

 

 

Theorem 2:  "Define,

 

{a,b,c,d} = {m+p, m-p, m+q, m-q}

{e,f,g,h}  = {m+r, m-r, m+s, m-s}

 

If,

 

p2+q2 = r2+s2    (eq.1)

p2+q2 = 4m2      (eq.2)

 

then 25(a4+b4+c4+d4-e4-f4-g4-h4)(a6+b6+c6+d6-e6-f6-g6-h6) = 21(a5+b5+c5+d5-e5-f5-g5-h5)2."

 

Also, for any m, then ak+bk+ck+dk = ek+fk+gk+hk,  for k = 1,2,3.  Hence it is a symmetric ideal solution (see Theorem 1 and 2) of the Prouhet-Tarry-Escott problem.  Note that eq.1 can be completely solved as {p,q,r,s} = {ux+vy, vx-uy, ux-vy, vx+uy} and substituting this into eq.2 yields,

 

(u2+v2)(x2+y2) = 4m2

 

which can be solved by using the formula for Pythagorean triples separately on {u,v} and {x,y}.

 

Theorem 3:  "Define,

 

{a,b,c,d,e,f} = {m+p, m-p, m+q, m-q, m+r, m-r}

{g,h,i,j,k,l}  =  {m+s, m-s, m+t, m-t, m+u, m-u}

 

If,

 

pk+qk+rk = sk+tk+uk,  for k = 2,4   (eq.1)

p2+q2+r2 = vm2                              (eq.2)

 

then, 49(a6+b6+c6+d6+e6+f6-g6-h6-i6-j6-k6-l6)(a8+b8+c8+d8+e8+f8-g8-h8-i8-j8-k8-l8) = (4/3)(v+21)(a7+b7+c7+d7+e7+f7-g7-h7-i7-j7-k7-l7)2."

 

And, for any m, an+bn+cn+dn+en+fn = gn+hn+in+jn+kn+ln,  for n = 1,2,3,4,5.  When v = 1, only two primitive solns are known with terms < 16,000,

 

{p,q,r,s,t,u} = {2, 289, 610, 170, 223, 614}  (by "Martin")

{p,q,r,s,t,u} = {1575, 11522, 11850, 4647, 8890, 13230}   (by James Waldby)

 

after this author posted the problem on the newsgroup sci.math.symbolic.  However, when v = 2, there are small solns starting with {p,q,r,s,t,u} = {3,5,8,0,7,7} and it can be shown there are an infinite number.  Let {p,q,r,s,t,u} = {z1+z2, -z1+z2, 2z2, z3+z4, -z3+z4, 2z4and eq.1 and eq.2 become,

 

z12+3z22 = z32+3z42      (eq.1)

2(z12+3z22) = 2m2        (eq.2)

 

Eq. 1 can be completely solved as {z1, z2, z3, z4} = {ux+3vy, vx-uy, ux-3vy, vx+uy} and eq.2 becomes,

 

(u2+3v2)(x2+3y2) = m2

 

and it is easy to solve this using a variant of the formula for Pythagorean triples separately on {u,v} and {x,y}.  The procedure can be extrapolated to higher systems.

 

 

III. Odd Powers

 

The difficulty with the system,

 

a1k + a2k + … + amk = b1k + b2k + … + bmk,  k = 1,3,5,…2n+1   (eq.1)

 

valid only for odd powers is that terms can be transposed or moved around to the form,

 

c1k + c2k + … + cpk = 0,  k = 1,3,5,…2n+1

 

and it is more challenging to find relations beyond k = 2n+1.  The trick then is to “fix the variables in place” by finding a balanced partition such that,

 

a) a1+a2+… +am = b1+b2+… +bm = 0,  or

b) eq.1 is valid for k = 2 as well

 

and hence terms can no longer be transposed arbitrarily.  (For the system k = 1,3,5, the two conditions are equivalent.)

 

a) Condition:  a1+a2+… +am = b1+b2+… +bm = 0.

 

Third powers:  “If ak+bk+ck = dk+ek+fk,  for k = 1,3, where a+b+c = d+e+f = 0, then 9abc(a6+b6+c6-d6-e6-f6) = 2(a9+b9+c9-d9-e9-f9).”

 

Fifth powers:  “If ak+bk+ck+dk = ek+fk+gk+hk,  for k = 1,3,5, where a+b+c+d = e+f+g+h = 0, (or equivalently, k = 1,2,3,5), then,

 

7(a4+b4+c4+d4-e4-f4-g4-h4)(a9+b9+c9+d9-e9-f9-g9-h9) = 12(a6+b6+c6+d6-e6-f6-g6-h6)(a7+b7+c7+d7-e7-f7-g7-h7).”

 

More succinctly, "Define Fk = ak+bk+ck+dk -(ek+fk+gk+hk).  If Fk = 0 for  k = {1,2,3,5}, then 7F4F9 = 12F6F7."

 

Note:  As these two systems have complete solns, given here (Form 13) and here (Form 5.3b), respectively, one can substitute those into the given identities and see that they are true.  For higher than fifth powers, there are only isolated results with this condition so it harder to come up with general statements.

 

b) Condition:  Valid for k = 2 as well.

 

As Bastien’s Theorem forbids a non-trivial result for the system k = 1,2,3 with only three terms on either side of the eqn, this condition yields a “general” result valid only for those with higher than third powers, starting with,

 

Define Fk: = x1k+x2k+x3k+x4k – (y1k+y2k+y3k+y4k).  If Fk = 0 for k = 1,2,3,5, then (9F7)(x12+x22+x32+x42) = 7F9.

 

Define Fk: = x1k+x2k+x3k+x4k+x5k – (y1k+y2k+y3k+y4k+y5k).  If Fk = 0 for k = 1,2,3,5,7, then (11F9)(x12+x22+x32+x42+x52) = 9F11.

 

Define Fk: = x1k+x2k+x3k+x4k+x5k+x6k – (y1k+y2k+y3k+y4k+y5k+y6k).  If Fk = 0 for k = 1,2,3,5,7,9, then (13F11)(x12+x22+x32+x42+x52+x62) = 11F13.

 

Note:  All known solns satisfy this and the form is easily extrapolated for higher systems.  However, it is proven only for fifth powers and a general proof is needed.

 

(Update, 12/21/09):  Jarek Wroblewski observed that the third conjecture can be generalized as (13F11)(x12+x22+...+x62+y12+y22+...+y62)/2 = 11F13, and similarly for the others, even if Fk is not zero for k = 2.  Of course, if  it is for k = 2, then it reduces to the form given above. 

 
  
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