[{"@context":"http:\/\/schema.org\/","@type":"BlogPosting","@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/problem-of-prouhet-tarry-escott-wikipedia\/#BlogPosting","mainEntityOfPage":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/problem-of-prouhet-tarry-escott-wikipedia\/","headline":"Problem of Prouhet-Tarry-Escott-Wikipedia","name":"Problem of Prouhet-Tarry-Escott-Wikipedia","description":"before-content-x4 A wikipedia article, free l’encyclop\u00e9i. after-content-x4 In mathematics, and more particularly in theory of numbers and in combination, the","datePublished":"2018-03-28","dateModified":"2018-03-28","author":{"@type":"Person","@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/author\/lordneo\/#Person","name":"lordneo","url":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/author\/lordneo\/","image":{"@type":"ImageObject","@id":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","url":"https:\/\/secure.gravatar.com\/avatar\/44a4cee54c4c053e967fe3e7d054edd4?s=96&d=mm&r=g","height":96,"width":96}},"publisher":{"@type":"Organization","name":"Enzyklop\u00e4die","logo":{"@type":"ImageObject","@id":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","url":"https:\/\/wiki.edu.vn\/wiki4\/wp-content\/uploads\/2023\/08\/download.jpg","width":600,"height":60}},"image":{"@type":"ImageObject","@id":"https:\/\/wikimedia.org\/api\/rest_v1\/media\/math\/render\/svg\/a601995d55609f2d9f5e233e36fbe9ea26011b3b","url":"https:\/\/wikimedia.org\/api\/rest_v1\/media\/math\/render\/svg\/a601995d55609f2d9f5e233e36fbe9ea26011b3b","height":"","width":""},"url":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/problem-of-prouhet-tarry-escott-wikipedia\/","wordCount":10923,"articleBody":"     (adsbygoogle = window.adsbygoogle || []).push({});before-content-x4A wikipedia article, free l’encyclop\u00e9i.        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4In mathematics, and more particularly in theory of numbers and in combination, the Problem of Prouhet-Tarry-Escott is to find, for each integer n {displaystyle n} , two sets        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4A {displaystyle A} And B {displaystyle B} of n {displaystyle n} whole each, such as:        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4\u2211a\u2208Aai= \u2211b\u2208Bbi{displaystyle sum _{ain A}a^{i}=sum _{bin B}b^{i}} for each i {displaystyle i} of first {Displaystyle 1} up to an integer k {displaystyle k} given [ first ] . And A {displaystyle A} And B {displaystyle B} Check these conditions, we write A = kB {displaystyle A=_{k}B} . We are looking for a size solution n {displaystyle n} minimum for a degree k {displaystyle k} given. This problem, still open, is named after Eug\u00e8ne Prouhet, who studied it in 1851, and Gaston Tarry and Edward Brind Escott, who considered him in the early 1910s. The greatest value of k {displaystyle k} for which we know a solution with n = k + first {displaystyle n=k+1} East k = 11 {displaystyle k=11} . A corresponding solution is given by the following sets [ 2 ] : A = { \u00b1 22 , \u00b1 sixty one , \u00b1 eighty six , \u00b1 127 , \u00b1 140 , \u00b1 151 }  , B = { \u00b1 35 , \u00b1 47 , \u00b1 ninety four , \u00b1 121 , \u00b1 146 , \u00b1 148 } {displaystyle A={pm 22,pm 61,pm 86,pm 127,pm 140,pm 151} ,qquad B={pm 35,pm 47,pm 94,pm 121,pm 146,pm 148}}  The whole k {displaystyle k} of the definition is the degree , and the whole n {displaystyle n} is here size  . It is easy to see that for any solution, we have k”>. We are therefore looking for a minimum size solution. For size n = 6 {displaystyle n=6} and the degree k = 5 {displaystyle k=5} , the two sets { 0 , 5 , 6 , 16 , 17 , 22 } {Displaystyle {0.5,6,16,17,22}}} And { first , 2 , ten , twelfth , 20 , 21 } {Displaystyle {1,2,10,12,20,21}} are a solution of the problem, since: 0 + 5 + 6 + 16 + 17 + 22 = 66 = first + 2 + ten + twelfth + 20 + 21 {Displaystyle 0+5+6+16+17+22 = 66 = 1+2+10+12+20+21} 02+ 52+ 62+ 162+ 172+ 222= 1090 = 12+ 22+ 102+ 122+ 202+ 212{Displaystyle 0^{2}+5^{2}+6^{2}+16^{2}+17^{2}+22^{2} = 1090 = 1^{2}+2^{2 }+10^{2}+12^{2}+20^{2}+21^{2}} 03+ 53+ 63+ 163+ 173+ 223= 19998 = 13+ 23+ 103+ 123+ 203+ 213{Displaystyle 0^{3}+5^{3}+6^{3}+16^{3}+17^{3}+22^{3} = 19998 = 1^{3}+2^{3 }+10^{3}+12^{3}+20^{3}+21^{3}} 04+ 54+ 64+ 164+ 174+ 224= 385234 = 14+ 24+ 104+ 124+ 204+ 214{Displaystyle 0^{4}+5^{4}+6^{4}+16^{4}+17^{4}+22^{4} = 385234 = 1^{4}+2^{4 }+10^{4}+12^{4}+20^{4}+21^{4}} 05+ 55+ 65+ 165+ 175+ 225= 7632966 = 15+ 25+ 105+ 125+ 205+ 215{Displaystyle 0^{5}+5^{5}+6^{5}+16^{5}+17^{5}+22^{5} = 7632966 = 1^{5}+2^{5 }+10^{5}+12^{5}+20^{5}+21^{5}} . A solution ideal is a solution whose size is equal to the degree + 1. The above solution is therefore ideal. In 1851, Eug\u00e8ne Prouhet posed the more general problem of distributing the whole x from 1 to n m in n classes, so that the sum of the powers k -Iems of the whole of each class is the same, for k = 0, 1, … the process he offers [ 3 ] amounts to number the classes from 0 to n – 1, to decompose each integer x – 1 in the number base n , to make the sum of its figures, to calculate the rest r of this sum Modulo n and to affect the whole x In the classroom r . In the case where n = 2, the placement of the whole x in one of the two clras of index 0 or 1 is done depending on whether the x -Thine term of the suite of Prouhet-Thue-Morse is 0 or 1. For example, the first 8 integers are divided into: 1, 4, 6, 7 on the one hand, and in 2, 3, 5, 8 D the other hand, and the sum of the powers k -Thine of these two classes coincides until k = 2. Leonard Eugene Dickson devotes a chapter of his History of numbers of numbers to ‘ Sets of integers with equal sums of like powers \u00bb [ 4 ] , and list no less than 70 articles on this subject. In his historical article [ 5 ] , Edward Maitland Wright notes that Prouhet’s article was only rediscovered until 1948. Recent developments are described by Peter Borwein and his co -authors [ 6 ] , [ 7 ] , [ 8 ] ; See also the article by Filaseta and Markovich [ 9 ] . A two -dimensional version was studied by Alpers and Tijdeman (2007). If the couple A = { a1, a2, … , an} {displaystyle A={a_{1},a_{2},ldots ,a_{n}}} And B = { b1, b2, … , bn} {displaystyle B={b_{1},b_{2},ldots ,b_{n}}} is a degree solution k {displaystyle k} , then for everything N \u2260 0 {displaystyle Nneq 0} and all M {displaystyle M} the couple A\u2032={Na1+M,Na2+M,\u2026,Nan+M}etB\u2032={Nb1+M,Nb2+M,\u2026,Nbn+M}{displaystyle A’={Na_{1}+M,Na_{2}+M,ldots ,Na_{n}+M}quad {rm {et}}quad B’={Nb_{1}+M,Nb_{2}+M,ldots ,Nb_{n}+M}} is still a solution of the same degree. Thus, the solution {0,5,6,16,17,22}=5{1,2,10,12,20,21}{Displaystyle {0.5,6,16,17,22} = _ {5} {1,2,10,12,20,21}}} also gives the solution {1,6,7,17,18,23}=5{2,3,11,13,21,22}.{Displaystyle {1,6,7,17,18,23} = _ {5} {2,3,11,13,21,22}.}. This observation makes it possible to normalize the solutions, by imposing for example that they contain only positive or zero integers, and that zero is there. We do not know any ideal solution for any degree, but we know [ 6 ] that for any degree k {displaystyle k} , there is a size solution n \u2264 k ( k + first ) \/2 + first {Displaystyle Nleq K (K+1)\/2+1} . Symmetrical solutions: a pair size solution n = 2 m {displaystyle n=2m} East symmetrical If each component is of the form {\u00b1c1,\u00b1c2,\u2026,\u00b1cm}.{displaystyle {pm c_{1},pm c_{2},ldots ,pm c_{m}}.} The solution given in the introduction is of this form. An unclean size solution is symmetrical If the components of the solution are opposed, that is to say A={a1,a2,\u2026,an}{displaystyle A={a_{1},a_{2},ldots ,a_{n}}} And B={\u2212a1,\u2212a2,\u2026,\u2212an}.{displaystyle B={-a_{1},-a_{2},ldots ,-a_{n}}.} Ideal and symmetrical solutions are known for degrees k \u2264 11 {displaystyle kleq 11} , except for k = ten {displaystyle k=10} [ ten ] : k = first {displaystyle k=1} { \u00b1 2 } =1{ \u00b1 first } {displaystyle {pm 2}=_{1}{pm 1}} k = 2 {displaystyle k=2} { – 2 , – first , 3 } =2{ 2 , first , – 3 } {Displaystyle {-2, -1.3} = _ {2} {2.1, -3}} k = 3 {displaystyle k=3} { \u00b1 3 , \u00b1 11 } =3{ \u00b1 7 , \u00b1 9 } {displaystyle {pm 3,pm 11}=_{3}{pm 7,pm 9}} k = 4 {displaystyle k=4} { – 8 , – 7 , first , 5 , 9 } =4{ 8 , 7 , – first , – 5 , – 9 } {Displaystyle {-8, -7,1,5,9} = _ {4} {8.7, -1, -1, -5, -9}} k = 5 {displaystyle k=5} { \u00b1 4 , \u00b1 9 , \u00b1 13 } =5{ \u00b1 first , \u00b1 11 , \u00b1 twelfth } {displaystyle {pm 4,pm 9,pm 13}=_{5}{pm 1,pm 11,pm 12}} k = 6 {displaystyle k=6} { – 51 , – 33 , – 24 , 7 , 13 , 38 , 50 } =6{ 51 , 33 , 24 , – 7 , – 13 , – 38 , – 50 } {Displaystyle {-51, -33, -24,7,13,38,50} = _ {6} {51,33.24, -7, -13, -38, -50}}} k = 7 {displaystyle k=7} { \u00b1 2 , \u00b1 16 , \u00b1 21 , \u00b1 25 } =7{ \u00b1 5 , \u00b1 14 , \u00b1 23 , \u00b1 24 } {displaystyle {pm 2,pm 16,pm 21,pm 25}=_{7}{pm 5,pm 14,pm 23,pm 24}} k = 8 {displaystyle k=8} { – 98 , – 82 , – 58 , – 34 , 13 , 16 , 69 , 75 , 99 } =8{ 98 , 82 , 58 , 34 , – 13 , – 16 , – 69 , – 75 , – 99 } {Displaystyle {-98, -82, -58, -34,13,6,69,75,99} = _ {8} {98.82.58.34, -13, -16, -69, -75, -75, -75, -75 , -99}} k = 9 {displaystyle k=9} { \u00b1 99 , \u00b1 100 , \u00b1 188 , \u00b1 301 , \u00b1 313 } =9{ \u00b1 71 , \u00b1 131 , \u00b1 180 , \u00b1 307 , \u00b1 308 } {displaystyle {pm 99,pm 100,pm 188,pm 301,pm 313}=_{9}{pm 71,pm 131,pm 180,pm 307,pm 308}} This last solution is given, with others, in Borwein et al. (2003). No ideal solution is known for k = ten {displaystyle k=10} . There is a more algebraic way of formulating the problem [ 11 ] : Proposition – The following conditions are equivalent: \u2211i=1naij=\u2211i=1nbij,(j=1,\u2026,k){displaystyle sum _{i=1}^{n}a_{i}^{j}=sum _{i=1}^{n}b_{i}^{j},quad (j=1,ldots ,k)} deg\u2061(\u220fi=1n(x\u2212ai)\u2212\u220fi=1n(x\u2212bi))\u2264n\u2212(k+1){displaystyle deg left(prod _{i=1}^{n}(x-a_{i})-prod _{i=1}^{n}(x-b_{i})right)leq n-(k+1)} (x\u22121)k+1|\u2211i=1nxai\u2212\u2211i=1nxbi.{displaystyle (x-1)^{k+1}left|sum _{i=1}^{n}x^{a_{i}}-sum _{i=1}^{n}x^{b_{i}}right..} Table of ContentsNotes [ modifier | Modifier and code ] References [ modifier | Modifier and code ] Related articles [ modifier | Modifier and code ] external links [ modifier | Modifier and code ] Notes [ modifier | Modifier and code ] \u2191 Borwein (2002), p. 85  \u2191 Solution given by Nutti Kuosa, Jean-Charles Meyrignac and Chen Shuwen, in 1999, see The Prouhet-Tarry-Escott problem .  \u2191 M. E. Prouhet, Memory on some relationships between the powers of numbers , C. R. Acad. Sci. Paris, s\u00e9rie I, vol. 33, 1851, p. 225 .  \u2191 (in) Leonard Eugene Dickson , History of the Theory of Numbers\u00a0 (in)  [Detail of editions] , vol. 2, 1919, chap. XXIV, p. 705-716 .  \u2191 Wright (1959)  \u2191 a et b Borwein et Ingalls (1944)  \u2191 Borwein (2002)  \u2191 Borwin, lison\u0115k it percival 2003  \u2191  (in) Michael Filaseta Et Maria Markovich , ‘ Newton polygons and the Prouhet\u2013Tarry\u2013Escott problem \u00bb , Journal of Number Theory , vol. 174, 2017 , p. 384\u2013400 (DOI\u00a0 10.1016\/J.Jt.2016.10.009 ) .  \u2191 Borwein (2002) and The Prouhet-Tarry-Escott problem .  \u2191 See Borwein and Ingalls (1944) for references.  References [ modifier | Modifier and code ] Related articles [ modifier | Modifier and code ] external links [ modifier | Modifier and code ]        (adsbygoogle = window.adsbygoogle || []).push({});after-content-x4"},{"@context":"http:\/\/schema.org\/","@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"item":{"@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/#breadcrumbitem","name":"Enzyklop\u00e4die"}},{"@type":"ListItem","position":2,"item":{"@id":"https:\/\/wiki.edu.vn\/all2en\/wiki32\/problem-of-prouhet-tarry-escott-wikipedia\/#breadcrumbitem","name":"Problem of Prouhet-Tarry-Escott-Wikipedia"}}]}]