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Prouhet–Tarry–Escott problem

In mathematics, the Prouhet–Tarry–Escott problem asks for two disjoint multisets A and B of n integers each, whose first k power sum symmetric polynomials are all equal. That is, the two multisets should satisfy the equations

for each integer i from 1 to a given k. It has been shown that n must be strictly greater than k. Solutions with are called ideal solutions. Ideal solutions are known for and for . No ideal solution is known for or for .[1]

This problem was named after Eugène Prouhet, who studied it in the early 1850s, and Gaston Tarry and Edward B. Escott, who studied it in the early 1910s. The problem originates from letters of Christian Goldbach and Leonhard Euler (1750/1751).

Examples edit

Ideal solutions edit

An ideal solution for n = 6 is given by the two sets { 0, 5, 6, 16, 17, 22 } and { 1, 2, 10, 12, 20, 21 }, because:

01 + 51 + 61 + 161 + 171 + 221 = 11 + 21 + 101 + 121 + 201 + 211
02 + 52 + 62 + 162 + 172 + 222 = 12 + 22 + 102 + 122 + 202 + 212
03 + 53 + 63 + 163 + 173 + 223 = 13 + 23 + 103 + 123 + 203 + 213
04 + 54 + 64 + 164 + 174 + 224 = 14 + 24 + 104 + 124 + 204 + 214
05 + 55 + 65 + 165 + 175 + 225 = 15 + 25 + 105 + 125 + 205 + 215.

For n = 12, an ideal solution is given by A = {±22, ±61, ±86, ±127, ±140, ±151} and B = {±35, ±47, ±94, ±121, ±146, ±148}.[2]

Other solutions edit

Prouhet used the Thue–Morse sequence to construct a solution with   for any  . Namely, partition the numbers from 0 to   into a) the numbers each with an even number of ones in its binary expansion and b) the numbers each with an odd number of ones in its binary expansion; then the two sets of the partition give a solution to the problem.[3] For instance, for   and  , Prouhet's solution is:

01 + 31 + 51 + 61 + 91 + 101 + 121 + 151 = 11 + 21 + 41 + 71 + 81 + 111 + 131 + 141
02 + 32 + 52 + 62 + 92 + 102 + 122 + 152 = 12 + 22 + 42 + 72 + 82 + 112 + 132 + 142
03 + 33 + 53 + 63 + 93 + 103 + 123 + 153 = 13 + 23 + 43 + 73 + 83 + 113 + 133 + 143.

Generalizations edit

A higher dimensional version of the Prouhet–Tarry–Escott problem has been introduced and studied by Andreas Alpers and Robert Tijdeman in 2007: Given parameters  , find two different multi-sets  ,   of points from   such that

 

for all   with   This problem is related to discrete tomography and also leads to special Prouhet-Tarry-Escott solutions over the Gaussian integers (though solutions to the Alpers-Tijdeman problem do not exhaust the Gaussian integer solutions to Prouhet-Tarry-Escott).

A solution for   and   is given, for instance, by:

  and
 .

No solutions for   with   are known.

See also edit

Notes edit

  1. ^ Borwein 2002, p. 85.
  2. ^ Solution found by Nuutti Kuosa, Jean-Charles Meyrignac and Chen Shuwen, in 1999.
  3. ^ Wright, E. M. (1959), "Prouhet's 1851 solution of the Tarry–Escott problem of 1910", The American Mathematical Monthly, 66: 199–201, doi:10.2307/2309513, MR 0104622.

References edit

  • Borwein, Peter B. (2002), "The Prouhet–Tarry–Escott problem", Computational Excursions in Analysis and Number Theory, CMS Books in Mathematics, Springer-Verlag, pp. 85–96, ISBN 0-387-95444-9, retrieved 2009-06-16 Chap.11.
  • Alpers, Andreas; Rob Tijdeman (2007), "The Two-Dimensional Prouhet-Tarry-Escott Problem" (PDF), Journal of Number Theory, 123 (2): 403–412, doi:10.1016/j.jnt.2006.07.001, retrieved 2015-04-01.

External links edit

prouhet, tarry, escott, problem, mathematics, asks, disjoint, multisets, integers, each, whose, first, power, symmetric, polynomials, equal, that, multisets, should, satisfy, equations, displaystyle, each, integer, from, given, been, shown, that, must, strictl. In mathematics the Prouhet Tarry Escott problem asks for two disjoint multisets A and B of n integers each whose first k power sum symmetric polynomials are all equal That is the two multisets should satisfy the equations a A a i b B b i displaystyle sum a in A a i sum b in B b i for each integer i from 1 to a given k It has been shown that n must be strictly greater than k Solutions with k n 1 displaystyle k n 1 are called ideal solutions Ideal solutions are known for 3 n 10 displaystyle 3 leq n leq 10 and for n 12 displaystyle n 12 No ideal solution is known for n 11 displaystyle n 11 or for n 13 displaystyle n geq 13 1 This problem was named after Eugene Prouhet who studied it in the early 1850s and Gaston Tarry and Edward B Escott who studied it in the early 1910s The problem originates from letters of Christian Goldbach and Leonhard Euler 1750 1751 Contents 1 Examples 1 1 Ideal solutions 1 2 Other solutions 2 Generalizations 3 See also 4 Notes 5 References 6 External linksExamples editIdeal solutions edit An ideal solution for n 6 is given by the two sets 0 5 6 16 17 22 and 1 2 10 12 20 21 because 01 51 61 161 171 221 11 21 101 121 201 211 02 52 62 162 172 222 12 22 102 122 202 212 03 53 63 163 173 223 13 23 103 123 203 213 04 54 64 164 174 224 14 24 104 124 204 214 05 55 65 165 175 225 15 25 105 125 205 215 For n 12 an ideal solution is given by A 22 61 86 127 140 151 and B 35 47 94 121 146 148 2 Other solutions edit Prouhet used the Thue Morse sequence to construct a solution with n 2 k displaystyle n 2 k nbsp for any k displaystyle k nbsp Namely partition the numbers from 0 to 2 k 1 1 displaystyle 2 k 1 1 nbsp into a the numbers each with an even number of ones in its binary expansion and b the numbers each with an odd number of ones in its binary expansion then the two sets of the partition give a solution to the problem 3 For instance for n 8 displaystyle n 8 nbsp and k 3 displaystyle k 3 nbsp Prouhet s solution is 01 31 51 61 91 101 121 151 11 21 41 71 81 111 131 141 02 32 52 62 92 102 122 152 12 22 42 72 82 112 132 142 03 33 53 63 93 103 123 153 13 23 43 73 83 113 133 143 Generalizations editA higher dimensional version of the Prouhet Tarry Escott problem has been introduced and studied by Andreas Alpers and Robert Tijdeman in 2007 Given parameters n k N displaystyle n k in mathbb N nbsp find two different multi sets x 1 y 1 x n y n displaystyle x 1 y 1 dots x n y n nbsp x 1 y 1 x n y n displaystyle x 1 y 1 dots x n y n nbsp of points from Z 2 displaystyle mathbb Z 2 nbsp such that i 1 n x i j y i d j i 1 n x i j y i d j displaystyle sum i 1 n x i j y i d j sum i 1 n x i j y i d j nbsp for all d j 0 k displaystyle d j in 0 dots k nbsp with j d displaystyle j leq d nbsp This problem is related to discrete tomography and also leads to special Prouhet Tarry Escott solutions over the Gaussian integers though solutions to the Alpers Tijdeman problem do not exhaust the Gaussian integer solutions to Prouhet Tarry Escott A solution for n 6 displaystyle n 6 nbsp and k 5 displaystyle k 5 nbsp is given for instance by x 1 y 1 x 6 y 6 2 1 1 3 3 6 6 7 7 5 5 2 displaystyle x 1 y 1 dots x 6 y 6 2 1 1 3 3 6 6 7 7 5 5 2 nbsp and x 1 y 1 x 6 y 6 1 2 2 5 5 7 7 6 6 3 3 1 displaystyle x 1 y 1 dots x 6 y 6 1 2 2 5 5 7 7 6 6 3 3 1 nbsp No solutions for n k 1 displaystyle n k 1 nbsp with k 6 displaystyle k geq 6 nbsp are known See also editEuler s sum of powers conjecture Beal s conjecture Jacobi Madden equation Lander Parkin and Selfridge conjecture Taxicab number Pythagorean quadruple Sums of powers a list of related conjectures and theorems Discrete tomographyNotes edit Borwein 2002 p 85 Solution found by Nuutti Kuosa Jean Charles Meyrignac and Chen Shuwen in 1999 Wright E M 1959 Prouhet s 1851 solution of the Tarry Escott problem of 1910 The American Mathematical Monthly 66 199 201 doi 10 2307 2309513 MR 0104622 References editBorwein Peter B 2002 The Prouhet Tarry Escott problem Computational Excursions in Analysis and Number Theory CMS Books in Mathematics Springer Verlag pp 85 96 ISBN 0 387 95444 9 retrieved 2009 06 16 Chap 11 Alpers Andreas Rob Tijdeman 2007 The Two Dimensional Prouhet Tarry Escott Problem PDF Journal of Number Theory 123 2 403 412 doi 10 1016 j jnt 2006 07 001 retrieved 2015 04 01 External links editWeisstein Eric W Prouhet Tarry Escott problem MathWorld Retrieved from https en wikipedia org w index php title Prouhet Tarry Escott problem amp oldid 1110256006, wikipedia, wiki, book, books, library,

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