019: Equal Sums of Like Powers and the Prouhet-Tarry-Escott (PTE) Problem

 
 

PART 8. Equals Sums of Like Powers and the PTE Problem

 
 I. Some General Conjectures and Problems
II. Some Theorems
III. Fifth Powers 

5.1    Four terms

5.2    Six terms

5.3    Seven terms

5.4    Eight terms

5.5    Ten terms

5.6    Twelve terms

 

Most of the identities I found for kth powers involve multigrade "Equal Sums of Like Powers (ESLP)" (multigrade being the case valid for several exponents).  For numerical examples that solve the "Prouhet-Tarry-Escott Problem (PTE)" in certain cases, see Chen Shuwen's site.  For the definitive site on the present status of ESLP, kindly see Jean-Charles Meyrignac's "Computing Minimal Equal Sums of Like Powers". 
 
  

I.  Some General Conjectures and Problems

 

1. Euler’s Extended Conjecture (EEC)

 

“The equation x1k + x2k +… + xmk = y1k + y2k +…+ ynk, (designated a [k,m,n]), has no solution in the integers for k > m+n, other than the trivial case when all xi = yi.”

 

This was formalized by R. Ekl’s 1998 “New Results in Equal Sums of Like Powers” which, as the title suggests, improved the known results on this problem.  Note that for [k,m,n] = [3,2,1], this partially reduces to Fermat's Last Theorem and is true as proven by A. Wiles, though this is the only proven result.  EEC is silent when k = m+n and whether this case has a solution or not is tackled by the stronger version,

 

2. Lander-Parkin-Selfridge Conjecture (LPS)

 

“The equation x1k + x2k +… + xmk = y1k + y2k +…+ ynk has a non-trivial solution in the integers for k > 3 if and only if m+n ≥ k.”

 

formalized by J. Meyrignac when he started EulerNet in 1999.  Thus this specifically demands that, for example, (4,1,3), (5,1,4), (6,1,5) have a soln.  While it is indeed true for the first two, for the third, no soln is known even for (6,1,6) so only time will tell if the conjecture will hold. 

 

3. Prouhet-Tarry-Escott Problem (PTE, for multi-grades)

 

“Does the system of equations x1k + x2k +… + xmk = y1k + y2k +…+ ymk,  for k = 1,2,…, n, have a non-trivial solution in the integers for any {m,n} where m > n?”

 

 

II. Some Theorems  (See also the section on Equal Sums of Like Powers in Assorted Identities.)

 

For simplicity, we will designate the multi-grade system a1k + a2k + …+ amk = b1k + b2k + …+ bmk,  valid for k =1,2,...,n, (call this system M) as,

 

[a1, a2,…, am] = [b1, b2,…, bm],  k = 1,2,...n

 

where m is its size and n is the degree.

 

Theorem 1Bastien’s Theorem  (L. Bastien, Sphinx-Oedipe, vol. 8, 1913, pp. 171-172.)

 

"For n > 1, if [a1, a2,…, am] = [b1, b2,…, bm] for k = 1,2,…n has a non-trivial soln, then m > n." 

 

In others words, for n >1, the size of the system is always greater than the degree.  To solve the system M up to exponent n, one has to use at least n+1 terms on one side of the equation (since Theorem 2 allows the other side to have just n terms).  The case m = n+1 is called an ideal solution and are now known for all degrees n  11 other than n = 10.  The next theorems allow transformations to be done on them.

 

Theorem 2Frolov’s Theorem  (M. Frolov, Bulletin de la Societe Mathematique de France, vol. 17, 1888, pp 69-83; vol. 20, 1892, pp.69-84.)

 

"If [a1, a2,…, am] = [b1, b2,…, bm], for k = 1,2,...n, then for any value x, it is true that [x+a1, x+a2,…, x+am] = [x+b1, x+b2,…, x+bm], for k = 1,2,...n."

   

For example, using the solution [1, 5, 6] = [2, 3, 7],  k = 1,2, then for any x,

 

(x+1)k + (x+5)k + (x+6)k = (x+2)k + (x+3)k + (x+7)k,  for k = 1,2  

 

Corollary 1: For any M, one can always set the first term a1 = 0 using the special case x = -a1.  For the example above, with x = -1 this becomes [0, 4, 5] = [1, 2, 6].

 

Corollary 2: One can also set the terms ai and bi such that their sum, Σai = Σbi = 0.  This is called the standard form by Escott and the importance of this form will be discussed in Theorem 6.  Ex,

 

[1, 5, 8, 12] = [2, 3, 10, 11],  k = 1,2,3

 

since (x+1)+(x+5)+(x+8)+(x+12) = 0 gives x = -26/4, then,

 

[-11, -3, 3, 11] = [-9, -7, 7, 9],  k = 1,2,3

 

after removing common factors.  It should be pointed out that not all become symmetric solutions.  For ex,

 

[1, 75, 87, 205] = [15, 37, 115, 201],  k = 1,2,3,  yields,

 

[-91, -17, -5, 113] = [-77, -55, 23, 109],  k = 1,2,3,5

 

Naturally enough, these are called non-symmetric solutions and why this example suddenly became applicable for k = 5 as well will be explained in Theorem 6.  The next three transformations generally double the number of terms but also extend the range of exponents k.

 

Theorem 3. Tarry-Escott Theorem  (Escott, Quarterly Journal of Mathematics, 1910, pp. 141-167; Tarry, L’Intermediaire des Mathematiciens, vol. 19, 1912, pp. 219-221.)

 

"If [a1, a2,…, am] = [b1, b2,…, bm], for k = 1,2,…n, then for any x, it is [a1,…, am, x+b1,…, x+bm] = [b1,…, bm, x+a1,…, x+am], for k = 1,2,…n+1."

 

Ex. [1, 5, 6] = [2, 3, 7],  k = 1,2 gives

 

[1, 5, 6, x+2, x+3, x+7] = [2, 3, 7, x+1, x+5, x+6],  k = 1,2,3

 

As expected, the number of terms doubles but with carefully chosen terms, this doubling can be prevented since some terms may appear on both sides of the equation and just cancel out.  For example, this one by Tarry,

 

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

 

using the theorem starting with x = 11 found the first ideal soln with degree n=7,

 

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

 

since eight terms on either side cancel out.  This author has found a generalization of this particular example to be discussed on the section on Sixth Powers. 

 

Theorem 4. (Starts with Odd Powers)

 

"If [a1, a2,…, am] = [b1, b2,…, bm],  for k = 1,3,…2n-1, then for any x, it is true that [x+a1,…, x+am, x-b1,…,x-bm] = [x-a1,…,x-am, x+b1,…, x+bm], for k = 1,2,3…2n."

 

Or alternatively, since we are dealing with odd powers and can move all terms to the LHS,

 

"If [a1, a2,…, ap] = 0, for k = 1,3,…2n-1, then [x+a1, x+a2…, x+ap] = [x-a1, x-a2,…,x-ap],  for k = 1,2,3…2n."

 

Ex.  [0, 7, 8, -1, -5, -9] = 0,  k = 1,3 gives,

 

[x+0, x+7, x+8, x-1, x-5, x-9] = [x-0, x-7, x-8, x+1, x+5, x+9],  k = 1,2,3,4

 

Note that appropriate pairs of terms taken from opposite sides of the equation have the common sum 2x.  If the original system has m = n+1 and is the special case a1 = 0, this also yields an ideal soln of degree 2n since the term x ± a1 is just the same (as one can see from the example above).  Cases with a1 = 0 are,

 

[0, 3] = [1, 2],  k = 1

[0, 7, 8] = [1, 5, 9],  k = 1,3

[0, 24, 33, 51] = [7, 13, 38, 50],  k = 1,3,5 

[0, 34, 58, 82, 98] = [13, 16, 69, 75, 99],  k = 1,3,5,7  (A. Letac, 1942)

 

with the last found pre-computer searching.  Parametric solns are known for all except the case k = 1,3,5,7.  (Along with a second soln, how did Letac find these back in 1942?)  It is then enough to find a k = 1,3,5,7,9 with a1= 0 to give an ideal soln for the missing degreee n =10 but, almost 70 years after Letac, none are known so far.

 

Theorem 5.  (Starts with Even Powers)

 

"If [a1, a2,…, am] = [b1, b2,…, bm], for k = 2,4,…2n, then for any x, it is [x+a1,…, x+am, x-a1,…, x-am] = [x+b1,…,x+bm, x-b1,…, x-bm],  for k = 1,2,3…2n+1".

 

Ex. [1, 9, 10] = [5, 6, 11],  k = 2,4 gives,

 

[x+1, x+9, x+10, x-1, x-9, x-10] = [x+5, x+6, x+11, x-5, x-6, x-11],  k = 1,2,3,4,5

 

If m = n+1, this automatically gives an ideal soln of degree 2n+1.  Note that pairs of terms now from the same side of the equation have the common sum 2x.  If x = 0, then odd power sums of terms on either side is Σai = Σbi = 0, a consequence implied by Frolov’s theorem.  Some examples with m = n+1 are,

 

[1, 7] = [5, 5],  k = 2 

[1, 9, 10] = [5, 6, 11],  k = 2,4

[2, 16, 21, 25] = [5, 14, 23, 24],  k = 2,4,6  (G. Tarry, 1913)

[71, 131, 180, 307, 308] = [99, 100, 188, 301, 313],  k = 2,4,6,8  (Borwein, Lisonek, Percival)

[22, 61, 86, 127, 140, 151] = [35, 47, 94, 121, 146, 148],  k = 2,4,6,8,10  (Kuosa, Meyrignac, and Shuwen)

 

Parameterizations are known for all these cases.  For k = 2,4,6,8, Letac earlier gave a method using an elliptic curve though it results in relatively large numbers.  For k = 2,4,6,8,10, after Choudhry and Wroblewski recently found also an elliptic curve (in 2008), it is now known that this system has an infinite number of solns.  In general, Theorems 4 and 5 yield what are called symmetric solutions since sums of appropriate pairs of terms have the common sum 2t.  While the above theorems apply to solns with any number of terms, given the special case that is ideal and non-symmetric, there is another that maintains the number of terms but still extends the range k,

 

Theorem 6. Gloden’s Theorem  (A. Gloden, Mehrgradige Gleichungen, Noordhoff, Groningen, 1944)

 

"If [a1, a2,…, am] = [b1, b2,…, bm],  k = 1,2,…n, where m = n+1, then [ma1-x, ma2-x,…, mam-x] = [mb1-x, mb2-x,…, mbm-x],  k = 1,2,…, n, n+2, where  x = a1+a2+…+am."

 

Gloden's theorem  is equivalent to reducing ideal solns to its standard form (when the sum of terms on each side is zero).  Note that, the sum is m(a1+a2+…+am )-mx and since x = a1+a2+…+am = b1+b2+…+bm, then both sides sum to zero.  Of course, one can also apply it to symmetric solns but it just gives a trivial result.  Example, using the non-symmetric ideal soln given earlier,

 

[1, 75, 87, 205] = [15, 37, 115, 201],  k = 1,2,3

 

we get,

 

[-91, -17, -5, 113] = [-77, -55, 23, 109],  k = 1,2,3,5 and for no other k,

 

where sum of terms Σai = Σbi = 0.  Thus, theorem 6 reduces ideal solns to standard form and, for the non-symmetric case, this reduction has the consequence of non-trivially increasing its range, though technically it is no longer ideal since it skips an exponent.

 

(Update, 8/7/09):  For an extensive collection of theorems involving equal sums of like powers, see Part 3 of Assorted Identities.

 

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