Prouhet–Tarry–Escott problem
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$${Published}$$ $${Online}$$ $${First}$$ $${(8/15/2024)}$$
$${Latest}$$ $${additions}$$ $${(8/15/2024)}$$
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0. Abstract【要約】
$${\rm \bf 0.~Abstract}$$
$${\small \rm In~mathematics,}$$
$${\small \rm the~Prouhet}$$-$${\small \rm Tarry}$$-$${\small \rm Escott~problem~asks~for}$$
$${\small \rm two~disjoint~multisets~A~and~B~of~n~integers~each,}$$
$${\small \rm whose~first~k~power~sum~symmetric~polynomials}$$
$${\small \rm are~all~equal.That~is,the~two~multisets}$$
$${\small \rm should~satisfy~the~equations}$$
$${\displaystyle \sum_{a\in A}a^i=\sum_{b\in B} b^i}$$
$${\small \rm for~each~integer~i~from~1~to~a~given~k.It~has~been}$$
$${\small \rm shown~that~n~must~be~strictly~greater~than~k.}$$
$${\small \rm Solutions~with~k=n-1~are~called~ideal~solutions.}$$
$${\small \rm Ideal~solutions~are~known~for~3\leq n\leq 10}$$
$${\small \rm and~for~n=12.}$$
$${\small \rm No~ideal~solution~is~known~for~n=11}$$
$${\small \rm or~for~n\geq 13.(unsolved~problem)}$$
$${\rm \bf 0. }$$要約
数学における,$${\rm Prouhet}$$-$${\rm Tarry}$$-$${\rm Escott}$$問題は,それぞれ$${n}$$個の整数の$${2}$$つの互いに素な多重集合$${A}$$と$${B}$$を要求し,その最初の$${k}$$乗和対称多項式はすべて等しい.
つまり,$${2}$$つの多重集合は方程式を満たす.
$${\displaystyle \sum_{a\in A}a^i=\sum_{b\in B} b^i}$$
各整数$${i}$$については$${1}$$から与えられた$${k}$$までの値.
$${n}$$は厳密に$${k}$$より大きいことが示されている.
$${\displaystyle k=n-1}$$の解が理想的な解と呼ばれています.
$${\displaystyle 3\leq n\leq 10}$$と$${\displaystyle n=12}$$に対して,理想的な解は知られています.
$${\displaystyle n=11}$$と$${\displaystyle n\geq 13}$$に対しては理想的な解は知られていない.$${(}$$未解決問題$${)}$$
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1. Introduction【序論】
$${\rm \bf 1.~Introduction}$$
$${\rm Explanation~of~words~and~symbols~used}$$
$${\rm this~paper.}$$
$${\rm \bf 1. }$$序論
この論文内で使用する用語と記号の説明.
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1.1 Problem name and problem for short【問題名と略称】
$${\small \rm \bf 1.1~Problem~name~and~problem~for~short}$$
$${\small \rm The~Prouhet}$$-$${\small \rm Tarry}$$-$${\small \rm Escott~problem,}$$
$${\small \rm or~\bf PTE~problem}$$ $${\small \rm for~short.}$$
$${\rm \bf 1. 1~}$$ 問題名と略称
$${\small \rm Prouhet}$$-$${\small \rm Tarry}$$-$${\small \rm Escott}$$の問題,又は
$${\small \rm \bf PTE}$$問題と略す.
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1.2 Solution notation【解の表記】
$${\small \rm \bf 1.2~Solution~notation}$$
$${\small \rm The~PTE~problem~can~be~stated~as:}$$
$${\small \rm Given~a~positive~integer~}$$$${n}$$,
$${\small \rm find~two~sets~of~integer~solutions}$$
$${\{ a_1, a_2, ... , a_m\}}$$$${\small \rm and}$$ $${\{ b_1, b_2, ... , b_m\}}$$
$${\small \rm such~that~the~integers~in~each~set~have}$$
$${\small \rm the~same~sum,the~same~sum~of~squares,~etc.,}$$
$${\small \rm up~to~and~including~the~same~sum~of}$$
$${n}$$$${\small \rm th~powers,~i.e.,}$$
$${\small \rm we~are~to~find~solutions~in~integers~of}$$
$${\small \rm the~system~of~equations.}$$
【$${\small \rm \bf Definition~1.2}$$】
$${\{ a_1^k+a_2^k+...+a_m^k~=~b_1^k + b_2^k + ... + b_m^k\}}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm Solutions~of~this~system~will~be~denoted~here}$$
$${\small \rm by~the~notation}$$
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm \bf 1.2}$$ 解の表記
$${\rm PTE}$$の問題は次の様に述べることができる.
正の整数$${n}$$が与えられると,$${2}$$組の整数解$${\{ a_1, a_2, ... , a_m\}}$$と$${\{ b_1, b_2, ... , b_m\}}$$を見つけ,
各セットの整数が同じ和,同じ平方和,同じ$${n}$$乗和などを持つ.
次の方程式系の整数解を見つける事と同じ.
【定義$${\small \rm \bf 1.2}$$】
$${\{ a_1^k+a_2^k+...+a_m^k~=~b_1^k + b_2^k + ... + b_m^k\}}$$
$${( k = 1, 2, ..., n )}$$
このシステムの解を次の様に表記する.
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 1, 2, ..., n )}$$
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1.3 Ideal solution【理想的な解】
【$${\small \rm \bf Definition~1.3}$$】$${\small \rm \bf Ideal~solution}$$
$${\small \rm Solutions~of~this~system}$$
$${a_1^k+a_2^k+...+a_m^k=b_1^k+b_2^k+...+b_m^k}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm with}$$ $${m = n + 1}$$ $${\small \rm are~called~ideal~solutions.}$$
【定義$${\small \rm \bf 1.3}$$】理想的な解
このシステムの解が
$${a_1^k+a_2^k+...+a_m^k=b_1^k+b_2^k+...+b_m^k}$$
$${( k = 1, 2, ..., n )}$$
$${m = n + 1}$$ の時,理想的な解と呼ぶ.
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1.4 Symmetric solutions【対称的な解】
【$${\small \rm \bf Definition~1.4}$$】$${\small \rm \bf Symmetric~solutions}$$
$${\small \rm Solutions~of~the~form}$$
$${[T+a_1,...,T+a_m,T-b_1,...,T-b_m]\\ \stackrel{\mathrm{k}}{=}[T+b_1,...,T+b_m,T-a_1,...,T-a_m]}$$
$${( k = 1, 2, ... , 2n )}$$ or
$${[T+a_1,...,T+a_m, T - a_1 , ... , T - a_m ]\\ \stackrel{\mathrm{k}}{=}[ T + b_1 , ... , T + b_m , T - b_1 , ... , T - b_m ]}$$
$${( k = 1, 2, ... , 2n + 1 )}$$
$${\small \rm are~called~symmetric~solutions.}$$
【定義$${\small \rm \bf 1.4}$$】対称的な解
解が次の型の時
$${[T+a_1,...,T+a_m,T-b_1,...,T-b_m]\\ \stackrel{\mathrm{k}}{=}[T+b_1,...,T+b_m,T-a_1,...,T-a_m]}$$
$${( k = 1, 2, ... , 2n )}$$ or
$${[T+a_1,...,T+a_m, T - a_1 , ... , T - a_m ]\\ \stackrel{\mathrm{k}}{=}[ T + b_1 , ... , T + b_m , T - b_1 , ... , T - b_m ]}$$
$${( k = 1, 2, ... , 2n + 1 )}$$
対称的な解と呼ぶ.
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1.5 Non-symmetric solutions【非対称的な解】
【$${\small \rm \bf Definition~1.5}$$】
$${\small \rm \bf Non}$$-$${\small \rm \bf symmetric~solutions}$$
$${\small \rm Otherwise, are~call~non}$$-$${\small \rm symmetric~solutions.}$$
【定義$${\small \rm \bf 1.5}$$】非対称的な解
それ以外は非対称的な解と呼ぶ.
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1.6 Ideal symmetric solutions【理想的で対称的な解】
【$${\small \rm \bf Definition~1.6}$$】
$${\small \rm \bf Ideal~symmetric~solutions}$$
$${\small \rm Ideal~and~symmetrical~solutions.}$$
【定義$${\small \rm \bf 1.6}$$】理想的で対称的な解
理想的で尚且つ対称的な解.
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1.7 Ideal non-symmetric solutions【理想的で非対称的な解】
【$${\small \rm \bf Definition~1.7}$$】
$${\small \rm \bf Ideal~non}$$-$${\small \rm \bf symmetric~solutions}$$
$${\small \rm Ideal~and~non}$$-$${\small \rm symmetric~solutions.}$$
【定義$${\small \rm \bf 1.7}$$】理想的で非対称的な解
理想的で尚且つ非対称的な解.
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1.8 Ideal prime solutions【理想的な素数解】
【$${\small \rm \bf Definition~1.8}$$】$${\small \rm \bf Ideal~prime~solutions}$$
$${\small \rm Ideal~and~prime~solutions.}$$
【定義$${\small \rm \bf 1.8}$$】理想的な素数解
理想的で尚且つ全て素数の解.
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1.9 Basic solutions【基本的な解】
【$${\small \rm \bf Definition~1.9}$$】$${\small \rm \bf Basic~solutions}$$
$${\small \rm Solutions~of~this~system}$$
$${a_1^k+a_2^k+...+a_m^k=b_1^k+b_2^k+...+b_m^k}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm with}$$ $${m > n + 1}$$ $${\small \rm are~called~basic~solutions.}$$
【定義$${\small \rm \bf 1.9}$$】基本的な解
このシステムの解が
$${a_1^k+a_2^k+...+a_m^k=b_1^k+b_2^k+...+b_m^k}$$
$${( k = 1, 2, ..., n )}$$
$${m > n + 1}$$の時を基本的な解と呼ぶ.
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1.10 Parametric solutions【パラメータ解】
【$${\small \rm \bf Definition~1.10}$$】$${\small \rm \bf Parametric~solutions}$$
$${\small \rm Solution~using~parameters.}$$
【定義$${\small \rm \bf 1.10}$$】パラメータ解
パラメータを用いた解.
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2. Historical flow【歴史的な流れ】
$${\small \rm \bf 2. Historical~flow}$$
$${\small \rm Does~not~cover~all~historical~trends.}$$
$${\small \rm \bf 2. }$$歴史的な流れ
全ての歴史的流れを網羅している訳ではない.
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2.1 【1750/1751】
$${\small \rm \bf 2.1}$$ 【$${\small \rm \bf 1750/1751}$$】
$${\small \rm The~problem~originates~from~letters~of}$$
$${\small \rm Christian~Goldbach~and~Leonhard~Euler.}$$
$${\small \rm \bf 2.1}$$ 【$${\small \rm 1750/1751}$$】
この問題は,クリスチャン･ゴールドバッハとレオンハルト･オイラーの手紙に由来しています.
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2.2 【early 1850s & early 1910s】
$${\small \rm \bf 2.2}$$【$${\small \rm \bf early~1850s}$$&$${\small \rm \bf early~1910s}$$】
$${\small \rm This~problem~was~named~after~Eugène~Prouhet, }$$
$${\small \rm who~studied~it~in~the~early~1850s,}$$
$${\small \rm and~Gaston~Tarry~and~Edward~B.Escott,}$$
$${\small \rm who~studied~it~in~the~early~1910s.}$$
$${\small \rm \bf 2.2}$$【$${1850}$$年代初頭&$${1910}$$年代初頭】
この問題は$${1850}$$年代初頭にそれを研究したウジェーヌ・プルーエと$${1910}$$年代初頭にそれを研究したガストン･タリーとエドワード･B･エスコットにちなんで名付けられました.
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2.3 【1851】
$${\small \rm \bf 2.3}$$【$${\small \rm \bf 1851}$$】
$${\small \rm Prouhet~used~the~Thue}$$-$${\small \rm Morse~sequence}$$
$${\small \rm to~construct~a~solution~with}$$$${n=2^{k}}$$ $${\small \rm for~any}$$ $${k}$$.
$${\small \rm Namely,partition~the~numbers~from~0~to}$$
$${2^{k+1}-1}$$ $${\small \rm into~A)~the~numbers~each~with}$$
$${\small \rm an~even~number~of~ones~in~its~binary~expansion}$$
$${\small \rm and~B)~the~numbers~each~with}$$
$${\small \rm an~odd~number~of~ones~in~its~binary~expansion;}$$
$${\small \rm then~the~two~sets~of~the~partition~give}$$
$${\small \rm a~solution~to~the~problem.}$$
$${\small \rm For~instance,~for}$$ $${n=8}$$ $${\small \rm and}$$ $${k=3}$$,
$${\small \rm Prouhet's~solution~is:}$$
$${\small 0^1+3^1 + 5^1 + 6^1 + 9^1 + 10^1 + 12^1 + 15^1 = 1^1 + 2^1 + 4^1 + 7^1 + 8^1 + 11^1 + 13^1 + 14^1}$$
$${\small 0^2 + 3^2 + 5^2 + 6^2 + 9^2 + 10^2 + 12^2 + 15^2 = 1^2 + 2^2 + 4^2 + 7^2 + 8^2 + 11^2 + 13^2 + 14^2}$$
$${\small 0^3 + 3^3 + 5^3 + 6^3 + 9^3 + 10^3 + 12^3 + 15^3 = 1^3 + 2^3 + 4^3 + 7^3 + 8^3 + 11^3 + 13^3 + 14^3}$$
$${\small \rm \bf 2.3}$$【$${\small \rm 1851}$$】
$${\small \rm Prouhet}$$は、$${\small \rm Thue}$$-$${\small \rm Morse}$$配列を使用して$${\displaystyle n=2^{k}}$$, 全ての $${\displaystyle k}$$に対して解を構築しました.
つまり,数字を$${0}$$から$${\displaystyle 2^{k+1}-1}$$までを分割する.
$${A)}$$ バイナリ展開に偶数を持つ数字
$${B)}$$ バイナリ展開に奇数を持つ数字
次に,$${2}$$つに分けたセットが問題の解を提供します.
例えば、$${\displaystyle n=8}$$ と$${\displaystyle k=3}$$
$${\small \rm Prouhet}$$の解は次のとおりです.
$${\small 0^1+3^1 + 5^1 + 6^1 + 9^1 + 10^1 + 12^1 + 15^1 = 1^1 + 2^1 + 4^1 + 7^1 + 8^1 + 11^1 + 13^1 + 14^1}$$
$${\small 0^2 + 3^2 + 5^2 + 6^2 + 9^2 + 10^2 + 12^2 + 15^2 = 1^2 + 2^2 + 4^2 + 7^2 + 8^2 + 11^2 + 13^2 + 14^2}$$
$${\small 0^3 + 3^3 + 5^3 + 6^3 + 9^3 + 10^3 + 12^3 + 15^3 = 1^3 + 2^3 + 4^3 + 7^3 + 8^3 + 11^3 + 13^3 + 14^3}$$
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2.4【Non-negative Integer Solutions of the case of k=1】
$${\small \rm \bf 2.4}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=1}$$】
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2.5【Non-negative Integer Solutions of the case of k=2】
$${\small \rm \bf 2.5}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=2}$$】
$${a_1^k+a_2^k+a_3^k=b_1^k+b_2^k+b_3^k}$$
$${k=1,2}$$
$${[a_1 , a_2 , a_3]\stackrel{\mathrm{3}}{=}[ b_1 , b_2 , b_3]}$$
$${\small \rm During~1750-1751,~Goldbach~and~Euler}$$
$${\small \rm ~had~studied~this~system.}$$
$${\small \rm The~general~solutions~are~easy~to~obtain.}$$
[ t + mn, t + pq, t + mp + nq ] = [ t + mp, t + nq, t + mn + pq ]
Solution chain of this system was obtained by A.Golden in 1940's.
[ 1, 17, 18 ] = [ 6, 7, 23 ] = [ 2, 13, 21 ] = [ 3, 11, 22 ]
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2.6【Non-negative Integer Solutions of the case of k=3】
$${\small \rm \bf 2.6}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=3}$$】
$${a_1^k+a_2^k+a_3^k+a_4^k=b_1^k+b_2^k+b_3^k+b_4^k}$$
$${k=1,2,3}$$
$${[a_1 , a_2 , ... , a_4 ]\stackrel{\mathrm{3}}{=}[ b_1 , b_2 , ... , b_4]}$$
L.E.Dickson gave the general solution for this system in 1910's.
Method for symmetric solution chain of this type was obtained by A.Golden in 1940's.
First known non-symmetric solution chain, by Chen Shuwen in 1997.
First known solution, symmetric prime solution, by Qiu Min, Wu Qiang in 2016.
Symmetric prime solution, by Chen Shuwen in 2016.
Non-symmetric prime solution, by Chen Shuwen in 2016.
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2.7【Non-negative Integer Solutions of the case of k=4】
$${\small \rm \bf 2.7}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=4}$$】
$${a_1^k+a_2^k+a_3^k+a_4^k+a_5^k=b_1^k+b_2^k+b_3^k+b_4^k+b_5^k}$$
$${k=1,2,3,4}$$
$${[a_1 , a_2 , ... , a_5]\stackrel{\mathrm{4}}{=}[ b_1 , b_2 , ... , b_5]}$$
J.Chernick gave a two-parameter solutions of this system in 1937.
All solutions by his method are symmetric.
J.L.Burchnall & T.W.Chaundy obtained a parameter solution for non-symmetric solution of this type in 1937.
First known non-symmetric solution, by J.L.Burchnall & T.W.Chaundy in 1937.
Ajai Choudhry obtained the complete symmetric solution and a parametric non-symmetric solution.of this type in 2000.
Symmetric prime solution, by Chen Shuwen in 2016.
Non-symmetric prime solution, by Chen Shuwen in 2016.
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2.8【Non-negative Integer Solutions of the case of k=5】
$${\small \rm \bf 2.8}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=5}$$】
$${a_1^k+a_2^k+a_3^k+a_4^k+a_5^k+a_6^k=b_1^k+b_2^k+b_3^k+b_4^k+b_5^k+b_6^k}$$
$${k=1,2,3,4,5}$$
$${[a_1 , a_2 , ... , a_6]\stackrel{\mathrm{5}}{=}[ b_1 , b_2 , ... , b_6]}$$
First known solution, by G.Tarry in 1912.
G.Tarry gave a two-parameter solution of this type in 1912.
A.Golden gave a four-parameter solution in 1912.[5]
Only one non-symmetric solution of this type was appeared in the literature.
It's obtained by A.Golden. However, Golden's method for this non-symmetric solution was unknown.
A.Golden gave a parameter method for solution chain of this type in 1940's.
Chen Shuwen found how to get non-symmetric solution of this type in 1995.
First known solution, symmetric prime solution, by Qiu Min, Wu Qiang in 2016.
Symmetric prime solution, by Chen Shuwen in 2016.
Non-symmetric prime solution, by Chen Shuwen in 2016.
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2.9【Non-negative Integer Solutions of the case of k=6】
$${\small \rm \bf 2.9}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=6}$$】
$${a_1^k+a_2^k+…+a_7^k=b_1^k+b_2^k+…+b_7^k.}$$
$${k=1,2,3,4,5,6}$$
$${[a_1 , a_2 , ... , a_7]\stackrel{\mathrm{6}}{=}[ b_1 , b_2 , ... , b_7]}$$
The first solution of this type was found out by E.B.Escott in 1910.
[ 0, 18, 27, 58, 64, 89, 101 ] = [ 1, 13, 38, 44, 75, 84, 102 ]
J.Chernick gave a two-parameter solution of this type in 1937.
No non-symmetric solution of this type is first found Chen Shuwen in 1997.
[ 0, 18, 19, 50, 56, 79, 81 ] = [ 1, 11, 30, 39, 68, 70, 84 ]
This one is also the smallest solution of this type that have been found out.
[ 0, 14, 43, 141, 156, 193, 199 ] = [ 3, 9, 46, 133, 175, 176, 204 ]
[ 0, 24, 31, 74, 106, 137, 147 ] = [ 4, 11, 52, 57, 119, 126, 150 ]
By Chen Shuwen, in 2001.
Each non-symmetric solution has its "co-symmetric" solution., For example
[ 0, 18, 19, 50, 56, 79, 81 ] = [ 1, 11, 30, 39, 68, 70, 84 ]
[ 0, 14, 16, 45, 54, 73, 83 ] = [ 3, 5, 28, 34, 65, 66, 84 ]
In fact, these two solutions are equivalent.
Smallest known solution, first known non-symmetric solution, by Chen Shuwen in 1997.
[ 0, 24, 31, 74, 106, 137, 147 ] = [ 4, 11, 52, 57, 119, 126, 150 ]
By Chen Shuwen in 2001.
[ 83, 191, 197, 383, 419, 557, 569 ] = [ 89, 149, 263, 317, 491, 503, 587 ]
First known solution, symmetric prime solution, by Qiu Min, Wu Qiang before Apr 2016.
Found independently by Chen Shuwen in Sep 2016.
[ 18443, 90263, 126173, 249863, 273803, 373553, 421433 ] = [ 22433, 70313, 170063, 194003, 317693, 353603, 425423 ]
Symmetric prime solution, by Chen Shuwen in 2016.
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2.10【Non-negative Integer Solutions of the case of k=7】
$${\small \rm \bf 2.10}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=7}$$】
$${a_1^k+a_2^k+…+a_8^k=b_1^k+b_2^k+…+b_8^k}$$
$${k=1,2,3,4,5,6,7}$$
$${[a_1 , a_2 , ... , a_8]\stackrel{\mathrm{7}}{=}[ b_1 , b_2 , ... , b_8]}$$
The first solution of this type was obtained by G.Tarry in 1913.
However, his method only can give one solution of this type.
[ 0, 4, 9, 23, 27, 41, 46, 50 ] = [ 1, 2, 11, 20, 30, 39 , 48, 49 ]
Crussol gave a parameter solution in 1913.
J.Chernick gave a two-parameter symmetric solution of this type in 1937.
Numerical examples by his method are
[ 0, 9, 10, 27, 41, 58, 59, 68 ] = [ 2, 3, 19, 20, 48, 49, 65, 66 ]
[ 0, 14, 19, 43, 57, 81, 86, 100 ] = [ 1, 9, 30, 32, 68, 70, 91, 99 ]
Non-symmetric solution of this type is first found by Chen Shuwen in 1997.
[ 0, 7, 23, 50, 53, 81, 82, 96 ] = [ 1, 5, 26, 42, 63, 72, 88, 95 ]
[ 0, 21, 82, 149, 155, 262, 278, 321 ] = [ 2, 17, 91, 126 , 174, 253, 285, 320 ]
[ 400033, 453073, 519373, 705013, 758053, 943693, 1009993, 1063033 ] = [ 413293, 426553, 545893, 665233, 797833, 917173, 1036513, 1049773 ]
Symmetric prime solution, by Chen Shuwen in 2016.
[ 10289, 14699, 27509, 41579, 42839, 65309, 68669, 77699 ] = [ 10709, 13859, 29399, 36749, 46829, 63419, 70139, 77489 ]
Non-symmetric prime solution, both by Chen Shuwen in 2016.
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2.11【Non-negative Integer Solutions of the case of k=8】
$${\small \rm \bf 2.11}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=8}$$】
$${a_1^k+a_2^k+…+a_9^k=b_1^k+b_2^k+…+b_9^k}$$
$${k=1,2,3,4,5,6,7,8}$$
$${[a_1 , a_2 , ... , a_9]\stackrel{\mathrm{8}}{=}[ b_1 , b_2 , ... , b_9]}$$
A.Latec gave two symmetric solutions in 1942.
[ 0, 24, 30, 83, 86, 133, 157, 181, 197 ] = [ 1, 17, 41, 65, 112, 115, 168, 174, 198 ]
[ 0, 26, 42, 124, 166, 237, 293, 335, 343 ] = [ 5, 13, 55, 111, 182, 224, 306, 322, 348 ]
Chen Shuwen made a computer search in 1997, and found that there is no other symmetric solution of this type in the range of max { ai, bi }< 400.
In 2002, by computer search, Peter Borwein, Petr Lisoněk, and Colin Percival proved that there is no new symmetric solution in the range of max { ai, bi }< 4000.
No any new result was obtained on this type during the pass 75 years.
In 2016, Chen Shuwen tried several methods for the new solution of this system, and found that
By full computer search, there is no new new symmetric solution in the range of max { ai, bi }< 13000.
By full computer search, there is no non-symmetric solution in the range of max { ai, bi }< 310.
By selective computer search, there is no new solution found.
Only one symmetric prime solution is found in the range of { ai, bi }< 100000000, by Chen Shuwen in 2016.
[ 3522263, 4441103, 5006543, 7904423, 9388703, 11897843, 13876883, 15361163, 15643883 ] = [ 3698963, 3981683, 5465963, 7445003, 9954143, 11438423, 14336303, 14901743, 15820583 ]
-------------------------------------------------
2.12【Non-negative Integer Solutions of the case of k=9】
$${\small \rm \bf 2.12}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=9}$$】
$${a_1^k+a_2^k+…+a_10^k=b_1^k+b_2^k+…+b_10^k}$$
$${k=1,2,3,4,5,6,7,8,9}$$
$${[a_1 , a_2 , ... , a_10]\stackrel{\mathrm{9}}{=}[ b_1 , b_2 , ... , b_10]}$$
Solution of this type was first found by A.Letac in 1940's.
[ 0, 3083, 3301, 11893, 23314, 24186, 35607, 44199, 44417, 47500 ] = [ 12, 2865, 3519, 11869, 23738, 23762, 35631, 43981, 44635, 47488 ]
The second solution was obtained by G.Palama in 1950 and C.J.Smyth in 1990.
Their methods both were based on Letac's method. See also ( h = 1, 2, 4, 6, 8 ) for more information.
The smallest two solutions are found by Peter Borwein, Petr Lisonek and Colin Percival in 2000.
[ 0, 12, 125, 213, 214, 412, 413, 501, 614, 626 ] = [ 5, 6, 133, 182, 242, 384, 444, 493, 620, 621 ]
[ 0, 63, 149, 326, 412, 618, 704, 881, 967, 1030 ] = [ 7, 44, 184, 270, 497, 533, 760, 846, 986, 1023 ]
The third smallest known solution is found by Chen Shuwen in 2023, based on the solution of ( k = 2, 4, 6, 8 ).
[ 0, 364, 810, 1229, 4310, 5344, 8425, 8844, 9290, 9654 ] = [ 40, 260, 1044, 1054, 4329, 5325, 8600, 8610, 9394, 9614 ]
Only one prime solution is found in the range of of { ai, bi }< 100000000, by Chen Shuwen in 2016.
[ 2589701, 2972741, 6579701, 9388661, 9420581, 15740741, 15772661, 18581621, 22188581, 22571621 ] = [ 2749301, 2781221, 6835061, 8399141, 10314341, 14846981, 16762181, 18326261, 22380101, 22412021 ]
-------------------------------------------------
2.13【Non-negative Integer Solutions of the case of k=11】
$${\small \rm \bf 2.13}$$【$${\small \rm \bf Non}$$-$${\small \rm \bf negative~Integer~Solutions}$$
$${\small \rm \bf of~the~case~of}$$$${\small \bf k=11}$$】
$${a_1^k+a_2^k+…+a_12^k=b_1^k+b_2^k+…+b_12^k}$$
$${k=1,2,3,4,5,6,7,8,9,10,11}$$
$${[a_1 , a_2 , ... , a_12]\stackrel{\mathrm{11}}{=}[ b_1 , b_2 , ... , b_12]}$$
Nuutti Kuosa discovered the following solution in 3 Sep1999, using a program written by Jean-Charles Meyrignac.
15110+14010+12710+8610+6110+2210=14810+14610+12110+9410+4710+3510
See Computing Minimal Equal Sums Of Like Powers for detail.
Chen Shuwen noticed in 7 Sep1999 that the above result is also a solution of ( k = 2, 4, 6, 8, 10 ) .
[ 22, 61, 86, 127, 140, 151 ] = [ 35, 47, 94, 121, 146, 148 ]
By using Theorem 4, Chen Shuwen translated Nuutti Kuosa & Jean-Charles Meyrignac's result into an ideal solution of ( k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ) .
[ 0, 11, 24, 65, 90, 129, 173, 212, 237, 278, 291, 302 ] = [ 3, 5, 30, 57, 104, 116, 186, 198, 245, 272, 297, 299 ]
This is a great progress on the Prouhet-Tarry-Escott problem.
David Broadhurst found the second solution of ( k = 2, 4, 6, 8, 10 ) in 2007.
In 2008, A.Choudhry and Jaroslaw Wroblewski proved that the system ( k = 2, 4, 6, 8, 10 ) has infinitely many solutions and give a method to produce new numerical solutions.
Prime solution of ( k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ) is first found by Jaroslaw Wroblewski in 2023.
[ 32058169621, 32367046651, 32732083141, 33883352071, 34585345321, 35680454791, 36915962911, 38011072381, 38713065631, 39864334561, 40229371051, 40538248081 ] = [ 32142408811, 32198568271, 32900561521, 33658714231,34978461541, 35315418301, 37280999401, 37617956161,38937703471, 39695856181, 40397849431, 40454008891 ]
Jaroslaw Wroblewski has found 10 prime solutions of this type.
See https://www.primepuzzles.net/problems/prob_084.htm for more information.
-------------------------------------------------
2.14【Non-negative Integer Solutions of the case of k=10,k≧12】
Unsolved problems.
-------------------------------------------------
2.0 REFERENCES【参考文献】
Copyright 1997-2023, Chen Shuwen
-------------------------------------------------
3. Known theorem【既知の定理】
$${\small \rm \bf 3.~Known~theorem}$$
$${\small \rm Introduction~to~some~known~theorems.}$$
$${\small \rm \bf 3.}$$既知の定理
いくつかの既知の定理の紹介.
-------------------------------------------------
3.1 Theorem 3.1 【定理 3.1】
$${\small \rm \bf Theorem 3.1}$$ [2][5]
$${\small \rm If}$$
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm then}$$
$${[ Sa_1+ T, Sa_2+ T, ... , Sa_m+ T ]\\~~~\stackrel{\mathrm{k}}{=}[ Sb_1+ T, Sb_2+ T, ... , Sb_m+ T ]}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm where}$$$${S}$$$${\small \rm and}$$$${T}$$ $${\small \rm are~arbitrary~integers.}$$
定理$${\small \rm \bf Theorem 3.1}$$ [2][5]
もし下記が解であれば
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 1, 2, ..., n )}$$
その時,次も解になる.
$${[ Sa_1+ T, Sa_2+ T, ... , Sa_m+ T ]\\~~~\stackrel{\mathrm{k}}{=}[ Sb_1+ T, Sb_2+ T, ... , Sb_m+ T ]}$$
$${( k = 1, 2, ..., n )}$$
ここで$${S}$$と$${T}$$は任意の整数.
-------------------------------------------------
3.2 Theorem 3.2 【定理 3.2】
$${\small \rm \bf Theorem 3.2}$$ [1][2]
$${\small \rm If}$$
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm then}$$
$${[ a_1 , ... , a_m , b_1+ T , ... , b_m + T ] \\~~~\stackrel{\mathrm{k}}{=}[ b_1 , ... , b_m , a_1 + T , ... , a_m + T ]}$$
$${( k = 1, 2, ... , n + 1 )}$$
$${\small \rm where}$$$${T}$$$${\small \rm is~arbitrary ~integer.}$$
定理$${\small \rm \bf Theorem 3.2}$$[1][2]
もし下記が解であれば
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 1, 2, ..., n )}$$
その時,次も解になる.
$${[ a_1 , ... , a_m , b_1+ T , ... , b_m + T ] \\~~~\stackrel{\mathrm{k}}{=}[ b_1 , ... , b_m , a_1 + T , ... , a_m + T ]}$$
$${( k = 1, 2, ... , n + 1 )}$$
ここで$${T}$$は任意の整数.
-------------------------------------------------
3.3 Theorem 3.3 【定理 3.3】
$${\small \rm \bf Theorem 3.3}$$ [5]
$${\small \rm If}$$
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 1, 3, ... , 2n - 1 )}$$
$${\small \rm then}$$
$${[ T + a_1 , ... , T + a_m , T - b_1 , ... , T - b_m ] \\ \stackrel{\mathrm{k}}{=}[ T + b_1 , ... , T + b_m , T - a_1 , ... , T - a_m ]}$$
$${( k = 1, 2, ... , 2n )}$$
$${\small \rm where}$$$${T}$$$${\small \rm is~arbitrary ~integer.}$$
定理$${\small \rm \bf Theorem 3.3}$$ [5]
もし下記が解であれば
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 1, 3, ... , 2n - 1 )}$$
その時,次も解になる.
$${[ T + a_1 , ... , T + a_m , T - b_1 , ... , T - b_m ] \\ \stackrel{\mathrm{k}}{=}[ T + b_1 , ... , T + b_m , T - a_1 , ... , T - a_m ]}$$
$${( k = 1, 2, ... , 2n )}$$
ここで$${T}$$は任意の整数.
-------------------------------------------------
3.4 Theorem 3.4 【定理 3.4】
$${\small \rm \bf Theorem 3.4}$$ [5]
$${\small \rm If}$$
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 2, 4, ... , 2n )}$$
$${\small \rm then}$$
$${[ T + a_1 , ... , T + a_m , T - a_1 , ... , T - a_m ]\\ \stackrel{\mathrm{k}}{=} [ T + b_1 , ... , T + b_m , T - b_1 , ... , T - b_m ]}$$
$${( k = 1, 2, ... , 2n + 1 )}$$
$${\small \rm where}$$$${T}$$$${\small \rm is~arbitrary ~integer.}$$
定理$${\small \rm \bf Theorem 3.4}$$ [5]
もし下記が解であれば
$${[a_1 , a_2 , ... , a_m ]\stackrel{\mathrm{k}}{=}[ b_1 , b_2 , ... , b_m]}$$
$${( k = 2, 4, ... , 2n )}$$
その時,次も解になる.
$${[ T + a_1 , ... , T + a_m , T - a_1 , ... , T - a_m ]\\ \stackrel{\mathrm{k}}{=} [ T + b_1 , ... , T + b_m , T - b_1 , ... , T - b_m ]}$$
$${( k = 1, 2, ... , 2n + 1 )}$$
ここで$${T}$$は任意の整数.
-------------------------------------------------
3.5 Theorem 3.5 【定理 3.5】
$${\small \rm \bf Theorem 3.5}$$ [5]
$${\small \rm If}$$
$${[a_1,a_2,...,a_{n+1}]\stackrel{\mathrm{k}}{=}[b_1,b_2,...,b_{n+1}]}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm then}$$
$${[ Sa_1 - T , Sa_2 - T , ..., Sa_{n+1} - T ]\\ \stackrel{\mathrm{k}}{=}[ Sb_1 - T , Sb_2 - T , ... , Sb_{n+1} - T ]}$$
$${( k = 1, 2, ... , n, n+2 )}$$
$${\small \rm where}$$ $${T = a_1+ a_2 + ... + a_{n+1},S = n + 1}$$.
定理$${\small \rm \bf Theorem 3.5}$$ [5]
もし下記が解であれば
$${[a_1,a_2,...,a_{n+1}]\stackrel{\mathrm{k}}{=}[b_1,b_2,...,b_{n+1}]}$$
$${( k = 1, 2, ..., n )}$$
その時,次も解になる.
$${[ Sa_1 - T , Sa_2 - T , ..., Sa_{n+1} - T ]\\ \stackrel{\mathrm{k}}{=}[ Sb_1 - T , Sb_2 - T , ... , Sb_{n+1} - T ]}$$
$${( k = 1, 2, ... , n, n+2 )}$$
ここで $${T = a_1+ a_2 + ... + a_{n+1},S = n + 1}$$.
-------------------------------------------------
3.6 Theorem 3.6 【定理 3.6】
$${\small \rm \bf Theorem 3.6}$$ [1][2]
$${\small \rm If~the~system}$$
$${a_1^k+a_2^k+...+a_m^k\\ ~~~\stackrel{\mathrm{k}}{=}b_1^k+b_2^k+...+b_m^k}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm have~a~non}$$-$${\small \rm trival~solution,~then}$$ $${m ≧ n +1}$$.
定理$${\small \rm \bf Theorem 3.6}$$ [1][2]
もしシステムが
$${a_1^k+a_2^k+...+a_m^k\\ ~~~\stackrel{\mathrm{k}}{=}b_1^k+b_2^k+...+b_m^k}$$
$${( k = 1, 2, ..., n )}$$
$${m ≧ n +1}$$の時,非自明な解を持っている.
-------------------------------------------------
3.0 REFERENCES【参考文献】
3.0 $${\rm REFERENCES}$$
[1] Hua L.K., Introduction to Number Theory, Springer, 1982.
[2] H.L.Dorwart and O.E.Brown, The Tarry-Escott Problem, Amer. Math. Monthly, 44(1937), 613-626.
[3] G.H.Hardy and E.M.Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford University Press, London, 1979, 328-339.
[4] E.Rees and C.Smyth, On the constant in the Tarry-Escott problem, Fifity Years of Poly- nomials, Lecture Notes in Math., vol. 1415, Springer, Berlin and New York, 1990, 196-208 MR91g:11030.
[5] A.Gloden, Mehrgradige Gleichungen, Noordhoff, Groningen, 1944.
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4. Basic solutions using the identity in the case of k=2,3,4
【k=2,3,4の恒等式を用いた基本的な解】
$${\small \rm \bf 4. Basic~solutions~using~the~identity~in~the}$$
$${\small \rm \bf case~of}$$ $${\small \bf k=2,3,4.}$$
$${\small \rm I~will~introduce~a~solution~that~uses~identities.}$$
$${\small \rm \bf 4.}$$ $${\small \bf k=2,3,4}$$ の場合の基本的な解の解法
恒等式を用いた基本的な解の解法を紹介する.
-------------------------------------------------
4.1 Solution using the identity in the case of k=2
【k=2 の場合の恒等式を用いた解法】
$${\small \rm \bf 4.1~~Solution~using~the~identity}$$
$${\small \rm \bf in~the~case~of}$$ $${\small \bf k=2}$$
$${\small \rm Let's~start~with~the~following~identity.}$$
$${\small \rm The~following~formula~holds~for~any}$$ $${a,b,c\in Z}$$
$${\small \rm and}$$ $${k=1,2}$$.
$${0^k+(a+b)^k+(b+c)^k+(c+a)^k\\=(a+b+c)^k+a^k+b^k+c^k}$$
$${\small \rm The~number~of~terms~on~the~left~and~right~sides}$$
$${\small \rm is~4.}$$
$${\small \rm If~you~actually~substitute}$$ $${k=1,2}$$ $${\small \rm and~calculate ~it.}$$
$${k=1→}$$
$${0^1+(a+b)^1+(b+c)^1+(c+a)^1\\=(a+b+c)^1+a^1+b^1+c^1\\=2a+2b+2c}$$
$${k=2→}$$
$${0^2+(a+b)^2+(b+c)^2+(c+a)^2\\=(a+b+c)^2+a^2+b^2+c^2\\=2(a^2+b^2+c^2+ab+bc+ca)}$$
$${\small \rm This~is~the~parameter~solution}$$
$${\small \rm of~the~basic~solution.}$$
$${\small \rm \bf 4.1}$$ $${\small \bf ~~k=2}$$ の場合の恒等式を用いた解法
次の恒等式から始めます.
$${k=1,2}$$ ,任意の$${a,b,c\in Z}$$に対して
$${0^k+(a+b)^k+(b+c)^k+(c+a)^k\\=(a+b+c)^k+a^k+b^k+c^k}$$
が成り立つ.
左辺と右辺の項数は$${4}$$個.
実際に$${k=1,2}$$を代入して計算してみると
$${k=1→}$$
$${0^1+(a+b)^1+(b+c)^1+(c+a)^1\\=(a+b+c)^1+a^1+b^1+c^1\\=2a+2b+2c}$$
$${k=2→}$$
$${0^2+(a+b)^2+(b+c)^2+(c+a)^2\\=(a+b+c)^2+a^2+b^2+c^2\\=2(a^2+b^2+c^2+ab+bc+ca)}$$
これが基本的な解のパラメータ解となる.
-------------------------------------------------
4.2 Solution using the identity in the case of k=3
【k=3 の場合の恒等式を用いた解法】
$${\small \rm \bf 4.2~~Solution~using~the~identity}$$
$${\small \rm \bf in~the~case~of}$$ $${\small \bf k=3}$$
$${\small \rm Let's~start~with~the~following~identity.}$$
$${\small \rm The~following~formula~holds~for~any}$$
$${a,b,c,d\in Z}$$ $${\small \rm and}$$$${k=1,2,3}$$.
$${\footnotesize 0^k+(a+b+c+d)^k+(a+b)^k+(b+c)^k+(c+d)^k+(a+c)^k+(a+d)^k+(b+d)^k\\=(a+b+c)^k+(b+c+d)^k+(c+d+a)^k+(d+a+b)^k+a^k+b^k+c^k+d^k}$$
$${\small \rm The~number~of~terms~on~the~left~and~right~sides}$$
$${\small \rm is~8.}$$
$${\small \rm If~you~actually~substitute}$$ $${k=1,2,3}$$ $${\small \rm and~calculate ~it.}$$
$${k=1→(\rm left)=(\rm right)}$$
$${~~~=4(a+b+c+d)}$$
$${k=2→(\rm left)=(\rm right)}$$
$${~~~=4(a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd)}$$
$${k=3→(\rm left)=(\rm right)}$$
$${2(a+b+c+d)(2a^2+2b^2+2c^2+2d^2+ab+ac+ad+bc+bd+cd)}$$
$${\small \rm This~is~the~parameter~solution}$$
$${\small \rm of~the~basic~solution.}$$
$${\small \rm \bf 4.2}$$ $${\small \bf ~~k=3}$$ の場合の恒等式を用いた解法
次の恒等式から始めます.
$${k=1,2,3}$$ ,任意の$${a,b,c,d\in Z}$$に対して
$${\footnotesize 0^k+(a+b+c+d)^k+(a+b)^k+(b+c)^k+(c+d)^k+(a+c)^k+(a+d)^k+(b+d)^k\\=(a+b+c)^k+(b+c+d)^k+(c+d+a)^k+(d+a+b)^k+a^k+b^k+c^k+d^k}$$
が成り立つ.
左辺と右辺の項数は$${8}$$個.
実際に$${k=1,2,3}$$を代入して計算してみると
$${k=1→}$$$${(}$$左辺$${)=(}$$右辺$${)\\~~~=4(a+b+c+d)}$$
$${k=2→}$$ $${(}$$左辺$${)=(}$$右辺$${)\\~~~=4(a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd)}$$
$${k=3→}$$ $${(}$$左辺$${)=(}$$右辺$${)\\~~~=2(a+b+c+d)(2a^2+2b^2+2c^2+2d^2+ab+ac+ad+bc+bd+cd)}$$
これが基本的な解のパラメータ解となる.
-------------------------------------------------
4.3 Solution using the identity in the case of k=4
【k=4 の場合の恒等式を用いた解法】
$${\small \rm \bf 4.3~~Solution~using~the~identity}$$
$${\small \rm \bf in~the~case~of}$$ $${\small \bf k=4}$$
$${\small \rm First, define}$$ $${(n,m)^k}$$.
$${\small \rm \bf Definition~4.3}$$
$${\small \rm From~the~set~containing}$$ $${n}$$ $${\small \rm elements,}$$
$${\small \rm choose~any}$$ $${m}$$, $${\small \rm and~in~all~cases,}$$
$${\small \rm the~sum~of~the}$$ $${m}$$ $${\small \rm elements~multitied~by}$$ $${k}$$
$${(n,m)^k}$$ $${\small \rm expressed~as}$$.
【$${\small \rm \bf Example~4.3}$$】
$${n=5,k=1,2,3,4}$$
$${\footnotesize (5,5)^k=(a+b+c+d+e)^k}$$
$${\footnotesize \begin{aligned} (5,4)^k &=(a+b+c+d)^k+(a+b+c+e)^k\\ &+(a+b+d+e)^k+(a+c+d+e)^k\\ &+(b+c+d+e)^k \end{aligned}}$$
$${\footnotesize \begin{aligned} (5,3)^k &=(a+b+c)^k+(a+b+d)^k\\ &+(a+b+e)^k+(a+c+d)^k\\ &+(a+c+e)^k+(a+d+e)^k\\ &+(b+c+d)^k+(b+c+e)^k\\&+(b+d+e)^k+(c+d+e)^k \end{aligned}}$$
$${\footnotesize \begin{aligned} (5,2)^k &=(a+b)^k+(a+c)^k+(a+d)^k\\&+(a+e)^k+(b+c)^k+(b+d)^k\\ &+(b+e)^k+(c+d)^k+(c+e)^k\\&+(d+e)^k \end{aligned}}$$
$${\footnotesize (5,1)^k=a^k+b^k+c^k+d^k+e^k}$$
$${\footnotesize (5,0)^k=0^k}$$
$${\small \rm then,~the~following~formula~is~established.}$$
$${k=1,2,3,4}$$
$${(5,5)^k+(5,3)^k+(5,1)^k\\=(5,4)^k+(5,2)^k+(5,0)^k}$$
$${\small \rm The~number~of~terms~on~the~left~and~right~sides}$$
$${\small \rm is~16.}$$
$${{}_5 \mathrm{C}_5+{}_5 \mathrm{C}_3+{}_5 \mathrm{C}_1={}_5 \mathrm{C}_4+{}_5 \mathrm{C}_2+{}_5 \mathrm{C}_0=16}$$
$${\small \rm This~is~the~parameter~solution}$$
$${\small \rm of~the~basic~solution.}$$
$${\small \rm \bf 4.3}$$ $${\small \bf ~~k=4}$$ の場合の恒等式を用いた解法
まず、$${(n,m)^k}$$を定義しておく.
【定義1】
$${n}$$個の要素を含む集合の中から,任意の$${m}$$個を選び,その全ての場合の$${m}$$個の要素の和をそれぞれ$${k}$$乗したものの総和を
$${(n,m)^k}$$と表す.
【例】$${n=5,k=1,2,3,4}$$
$${\footnotesize (5,5)^k=(a+b+c+d+e)^k}$$
$${\footnotesize \begin{aligned} (5,4)^k &=(a+b+c+d)^k+(a+b+c+e)^k\\ &+(a+b+d+e)^k+(a+c+d+e)^k\\ &+(b+c+d+e)^k \end{aligned}}$$
$${\footnotesize \begin{aligned} (5,3)^k &=(a+b+c)^k+(a+b+d)^k\\ &+(a+b+e)^k+(a+c+d)^k\\ &+(a+c+e)^k+(a+d+e)^k\\ &+(b+c+d)^k+(b+c+e)^k\\&+(b+d+e)^k+(c+d+e)^k \end{aligned}}$$
$${\footnotesize \begin{aligned} (5,2)^k &=(a+b)^k+(a+c)^k+(a+d)^k\\&+(a+e)^k+(b+c)^k+(b+d)^k\\ &+(b+e)^k+(c+d)^k+(c+e)^k\\&+(d+e)^k \end{aligned}}$$
$${\footnotesize (5,1)^k=a^k+b^k+c^k+d^k+e^k}$$
$${\footnotesize (5,0)^k=0^k}$$
この時,次の式が成り立つ.
$${k=1,2,3,4}$$
$${(5,5)^k+(5,3)^k+(5,1)^k\\=(5,4)^k+(5,2)^k+(5,0)^k}$$
ここで左辺と右辺の項数は$${16}$$個.
$${{}_5 \mathrm{C}_5+{}_5 \mathrm{C}_3+{}_5 \mathrm{C}_1={}_5 \mathrm{C}_4+{}_5 \mathrm{C}_2+{}_5 \mathrm{C}_0=16}$$
これが$${\small \bf k=4}$$の基本的な解のパラメータ解となる.
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5. The solution of the basic solutions for all k≧2
【全ての k≧2 に対する基本的な解の解法】
$${\small \rm \bf 5.~~The~solution~of~the~basic~solutions}$$
$${\small \rm \bf for~all}$$$${\small \bf k≧2}$$.
$${\small \rm \bf 5. }$$全ての$${\small \bf k≧2}$$に対する基本的な解の解法.
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5.1 General solution of the basic solutions of all k≧2
【k≧2の基本的な解の一般解】
$${\small \rm \bf 5.1~~General~solution~of~the~basic~solutions}$$
$${\small \rm \bf of~all}$$ $${\bf k≧2}$$
$${\small \rm Here,~we~will~display~the~basic~solution~of~all}$$ $${k≧2}$$.
$${\small \rm The~solution~shown~by~Prouhet~is~a~solution}$$
$${\small \rm that~divides~everything~from}$$ $${0}$$ $${\small \rm to}$$ $${2^{k+1}-1}$$ $${\small \rm into~a~set~of}$$ $${2}$$.
$${\small \rm Here~is~another~solution.}$$
$${\small \rm \bf Theorem~5.1}$$
$${\small \rm A~solution~divided~into~two~sets~consisting~of}$$
$${\small \rm a~sum~set~represented~by}$$ $${0}$$ $${\small \rm and~any}$$ $${k+1}$$ $${\small \rm parameters.}$$
$${\small \rm If}$$ $${k=n-1}$$ $${\small \rm is~an~odd~number.}$$
$${\footnotesize (n,n)^k+(n,n-2)^k+(n,n-4)^k+…+(n,0)^k\\=(n,n-1)^k+(n,n-3)^k+…+(n,1)^k}$$
$${\small \rm If}$$ $${k=n-1}$$ $${\small \rm is~an~even~number.}$$
$${\footnotesize (n,n)^k+(n,n-2)^k+(n,n-4)^k+…+(n,1)^k\\=(n,n-1)^k+(n,n-3)^k+…+(n,0)^k}$$
$${\small \rm The~number~of~each~term~is}$$
$${\small \rm If}$$ $${k=n-1}$$ $${\small \rm is~an~odd~number.}$$
$${\footnotesize {}_n \mathrm{C}_n+{}_n \mathrm{C}_{n-2} +…+{}_n \mathrm{C}_0 \\={}_n \mathrm{C}_{n-1} +{}_n \mathrm{C}_{n-3} +…+{}_n \mathrm{C}_1 =2^n}$$
$${\small \rm If}$$ $${k=n-1}$$ $${\small \rm is~an~even~number.}$$
$${\footnotesize {}_n \mathrm{C}_n+{}_n \mathrm{C}_{n-2} +…+{}_n \mathrm{C}_1 \\={}_n \mathrm{C}_{n-1} +{}_n \mathrm{C}_{n-3} +…+{}_n \mathrm{C}_0 =2^n}$$
$${\small \rm \bf 5.1}$$ 全ての$${\small \bf ~~k≧2}$$ に対する基本的な解の一般解
ここで全ての$${k≧2}$$の基本的解の表示をします.
$${\small \rm Prouhet}$$が示した解は$${0}$$から$${\displaystyle 2^{k+1}-1}$$までの全てを$${2}$$個の集合に分けた解である.
ここでは別の解を示す.
定理 $${\small \rm \bf 5.1}$$
$${0}$$と任意の$${\displaystyle k+1}$$個のパラメータで表される和集合で構成した2個の集合に分けた基本的な解のパラメータ解は,全ての$${k≧2}$$ ,$${k=n-1}$$に対して,
次の恒等式(基本的な解)が成り立つ.
$${k=n-1}$$が奇数の場合
$${\footnotesize (n,n)^k+(n,n-2)^k+(n,n-4)^k+…+(n,0)^k\\=(n,n-1)^k+(n,n-3)^k+…+(n,1)^k}$$
$${k=n-1}$$が偶数の場合
$${\footnotesize (n,n)^k+(n,n-2)^k+(n,n-4)^k+…+(n,1)^k\\=(n,n-1)^k+(n,n-3)^k+…+(n,0)^k}$$
それぞれの項数は
$${k=n-1}$$が奇数の場合
$${\footnotesize {}_n \mathrm{C}_n+{}_n \mathrm{C}_{n-2} +…+{}_n \mathrm{C}_0 \\={}_n \mathrm{C}_{n-1} +{}_n \mathrm{C}_{n-3} +…+{}_n \mathrm{C}_1 =2^n}$$
$${k=n-1}$$が偶数の場合
$${\footnotesize {}_n \mathrm{C}_n+{}_n \mathrm{C}_{n-2} +…+{}_n \mathrm{C}_1 \\={}_n \mathrm{C}_{n-1} +{}_n \mathrm{C}_{n-3} +…+{}_n \mathrm{C}_0 =2^n}$$
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5.2 Specific numerical display of basic solutions of k≧2
【k≧2の基本的解の具体的数値表示】
$${\small \rm \bf 5.2~~Specific~numerical~display~of}$$
$${\small \rm \bf basic~solutions~of}$$$${\bf k≧2}$$
$${\small \rm \bf Theorem~5.2}$$
$${\small \rm A~solution~that~divides~all~numbers~from}$$
$${0}$$ $${\small \rm to}$$ $${\displaystyle k+1}$$ $${\small \rm into~even~and~odd~numbers.}$$
$${\small \rm (Allow~duplication)}$$
$${k≧2}$$ $${\small \rm , when~all~elements~are}$$ $${1}$$
$${(n,m)^k={}_n \mathrm{C}_m m^k}$$
$${\small \rm Since~it~will~be}$$, $${\small \rm from~[Theorem~5.1]}$$
$${\small \rm The~basic~solution~is~expressed~in~the~following.}$$
$${\small \rm If}$$ $${k=n-1}$$ $${\small \rm is~an~odd~number.}$$
$${\footnotesize {}_n \mathrm{C}_n n^k+{}_n \mathrm{C}_{n-2} (n-2)^k+…+{}_n \mathrm{C}_0 0^k\\={}_n \mathrm{C}_{n-1} (n-1)^k+{}_n \mathrm{C}_{n-3} (n-3)^k+…+{}_n \mathrm{C}_1 1^k}$$
$${\small \rm If}$$ $${k=n-1}$$ $${\small \rm is~an~even~number.}$$
$${\footnotesize {}_n \mathrm{C}_n n^k+{}_n \mathrm{C}_{n-2} (n-2)^k+…+{}_n \mathrm{C}_1 1^k\\={}_n \mathrm{C}_{n-1} (n-1)^k+{}_n \mathrm{C}_{n-3} (n-3)^k+…+{}_n \mathrm{C}_0 0^k}$$
$${\small \rm The~number~of~each~term~is}$$
$${\small \rm If}$$ $${k=n-1}$$ $${\small \rm is~an~odd~number.}$$
$${\footnotesize {}_n \mathrm{C}_n+{}_n \mathrm{C}_{n-2} +…+{}_n \mathrm{C}_0 \\={}_n \mathrm{C}_{n-1} +{}_n \mathrm{C}_{n-3} +…+{}_n \mathrm{C}_1 =2^n}$$
$${\small \rm If}$$ $${k=n-1}$$ $${\small \rm is~an~even~number.}$$
$${\footnotesize {}_n \mathrm{C}_n+{}_n \mathrm{C}_{n-2} +…+{}_n \mathrm{C}_1 \\={}_n \mathrm{C}_{n-1} +{}_n \mathrm{C}_{n-3} +…+{}_n \mathrm{C}_0 =2^n}$$
$${\small \rm \bf 5.2~~}$$$${\bf k≧2}$$の基本的解の具体的数値表示
定理 $${\small \rm \bf 5.2}$$
$${0}$$から$${\displaystyle k+1}$$までの全ての数を偶数と奇数に分けた解.(重複を許す)
$${k≧2}$$ ,全ての要素を$${1}$$とした時
$${(n,m)^k={}_n \mathrm{C}_m m^k}$$となるので,【定理 5.1】より
基本的な解は次で表される.
$${k=n-1}$$が奇数の場合
$${\footnotesize {}_n \mathrm{C}_n n^k+{}_n \mathrm{C}_{n-2} (n-2)^k+…+{}_n \mathrm{C}_0 0^k\\={}_n \mathrm{C}_{n-1} (n-1)^k+{}_n \mathrm{C}_{n-3} (n-3)^k+…+{}_n \mathrm{C}_1 1^k}$$
$${k=n-1}$$が偶数の場合
$${\footnotesize {}_n \mathrm{C}_n n^k+{}_n \mathrm{C}_{n-2} (n-2)^k+…+{}_n \mathrm{C}_1 1^k\\={}_n \mathrm{C}_{n-1} (n-1)^k+{}_n \mathrm{C}_{n-3} (n-3)^k+…+{}_n \mathrm{C}_0 0^k}$$
それぞれの項数は
$${k=n-1}$$が奇数の場合
$${\footnotesize {}_n \mathrm{C}_n+{}_n \mathrm{C}_{n-2} +…+{}_n \mathrm{C}_0 \\={}_n \mathrm{C}_{n-1} +{}_n \mathrm{C}_{n-3} +…+{}_n \mathrm{C}_1 =2^n}$$
$${k=n-1}$$が偶数の場合
$${\footnotesize {}_n \mathrm{C}_n+{}_n \mathrm{C}_{n-2} +…+{}_n \mathrm{C}_1 \\={}_n \mathrm{C}_{n-1} +{}_n \mathrm{C}_{n-3} +…+{}_n \mathrm{C}_0 =2^n}$$
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5.3 Specific example of the case of k=6
【k=6 の場合の具体例】
$${\small \rm \bf 5.3~~Specific~example~of~the~case~of}$$ $${\bf k=6}$$
$${\small \rm In~the~case~of}$$ $${k=n-1=6}$$, $${\small \rm the~following~expression~holds~from~[Theorem 2]}$$
$${ {}_7 \mathrm{C}_7 7^k+{}_7 \mathrm{C}_5 5^k+{}_7 \mathrm{C}_3 3^k+{}_7 \mathrm{C}_1 1^k\\={}_7 \mathrm{C}_6 6^k+{}_7 \mathrm{C}_4 4^k+{}_n \mathrm{C}_2 2^k+{}_n \mathrm{C}_0 0^k}$$
$${\small \rm then}$$
$${{}_7 \mathrm{C}_7 ={}_7 \mathrm{C}_0=1\\{}_7 \mathrm{C}_6 ={}_7 \mathrm{C}_1=7\\{}_7 \mathrm{C}_5 ={}_7 \mathrm{C}_2=21\\{}_7 \mathrm{C}_4 ={}_7 \mathrm{C}_3=35}$$
$${\small \rm That~is}$$
$${1･7^k+21･5^k+35･3^k+7･1^k\\=7･6^k+35･4^k+21･2^k+1･0^k}$$
$${\small \rm All~of}$$ $${k=1,2,3,4,5,6,}$$.
$${k=1→}$$$${\small \rm (Left)=(Right)=224}$$
$${k=2→}$$$${\small \rm (Left)=(Right)=896}$$
$${k=3→}$$$${\small \rm (Left)=(Right)=3920}$$
$${k=4→}$$$${\small \rm (Left)=(Right)=18368}$$
$${k=5→}$$$${\small \rm (Left)=(Right)=90944}$$
$${k=6→}$$$${\small \rm (Left)=(Right)=471296}$$
$${\small \rm \bf 5.3~~}$$ $${\bf k=6}$$の場合の具体例
$${k=n-1=6}$$の時【定理 2】より次式が成り立つ
$${ {}_7 \mathrm{C}_7 7^k+{}_7 \mathrm{C}_5 5^k+{}_7 \mathrm{C}_3 3^k+{}_7 \mathrm{C}_1 1^k\\={}_7 \mathrm{C}_6 6^k+{}_7 \mathrm{C}_4 4^k+{}_n \mathrm{C}_2 2^k+{}_n \mathrm{C}_0 0^k}$$
となる.ここで
$${{}_7 \mathrm{C}_7 ={}_7 \mathrm{C}_0=1\\{}_7 \mathrm{C}_6 ={}_7 \mathrm{C}_1=7\\{}_7 \mathrm{C}_5 ={}_7 \mathrm{C}_2=21\\{}_7 \mathrm{C}_4 ={}_7 \mathrm{C}_3=35}$$
となるので,結局次の基本的な解が求まる.
$${1･7^k+21･5^k+35･3^k+7･1^k\\=7･6^k+35･4^k+21･2^k+1･0^k}$$
$${k=1,2,3,4,5,6,}$$の全てで成り立つ.
$${k=1→}$$$${(}$$左辺$${)=(}$$右辺$${)=224}$$
$${k=2→}$$$${(}$$左辺$${)=(}$$右辺$${)=896}$$
$${k=3→}$$$${(}$$左辺$${)=(}$$右辺$${)=3920}$$
$${k=4→}$$$${(}$$左辺$${)=(}$$右辺$${)=18368}$$
$${k=5→}$$$${(}$$左辺$${)=(}$$右辺$${)=90944}$$
$${k=6→}$$$${(}$$左辺$${)=(}$$右辺$${)=471296}$$
項数は左辺$${=}$$右辺$${=2^6=64}$$個
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6. The solution of the ideal solution using the identity【恒等式を用いた理想的な解の解法】
$${\small \rm \bf 6.~~The~solution~of~the~ideal~solutions}$$
$${\small \rm \bf using~the~identity}$$
【$${\small \rm \bf Definition~1.3}$$】$${\small \rm \bf Ideal~solution}$$
$${\small \rm Solutions~of~this~system}$$
$${a_1^k+a_2^k+...+a_m^k=b_1^k+b_2^k+...+b_m^k}$$
$${( k = 1, 2, ..., n )}$$
$${\small \rm with}$$ $${m = n + 1}$$ $${\small \rm are~called~ideal~solutions.}$$
$${\small \rm \bf 6.}$$理想的な解の解法
【定義$${\small \rm \bf 1.3}$$】理想的な解
このシステムの解が
$${a_1^k+a_2^k+...+a_m^k=b_1^k+b_2^k+...+b_m^k}$$
$${( k = 1, 2, ..., n )}$$
$${m = n + 1}$$ の時,理想的な解と呼ぶ.
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6.1 The ideal parameter solution in the case of k=2
【k=2の場合の理想的なパラメータ解】
$${\small \rm \bf 6.1~~The~ideal~parameter~solution~in~the}$$
$${\small \rm \bf case~of}$$ $${\bf k=2}$$
$${\small \rm If~you~substitute}$$ $${c=a+b}$$ $${\small \rm to~the~identity}$$
$${\small \rm of~the~basic~solution~of~the~previous}$$ $${k=2}$$
$${\small \rm and~reduce~the~common~term~of}$$
$${\small \rm the~left~right~side, it~will~be~as~follows.}$$
$${\scriptsize 0^k+(a+2b)^k+(2a+b)^k=(2a+2b)^k+a^k+b^k}$$
$${k=1→}$$
$${\footnotesize 0^1+(a+2b)^1+(2a+b)^1=(2a+2b)^1+a^1+b^1\\=3(a+b)}$$
$${k=2→}$$
$${\footnotesize 0^2+(a+2b)^2+(2a+b)^2=(2a+2b)^2+a^2+b^2\\=5a^2+8ab+5b^2}$$
$${\small \rm Because~this~makes~the~number~of~terms}$$
$${\small \rm on~the~left~and~right~sides}$$ $${3}$$,
$${\small \rm It~is~a~parameter~solution~of~the~ideal~solution.}$$
$${\small \rm \bf 6.1~~}$$ $${\bf k=2}$$の場合の理想的なパラメータ解
先の$${k=2}$$の基本的解の恒等式に$${c=a+b}$$を代入して左辺右辺の共通項を減らすと次の様になる.
$${\scriptsize 0^k+(a+2b)^k+(2a+b)^k=(2a+2b)^k+a^k+b^k}$$
$${k=1→}$$
$${\footnotesize 0^1+(a+2b)^1+(2a+b)^1=(2a+2b)^1+a^1+b^1\\=3(a+b)}$$
$${k=2→}$$
$${\footnotesize 0^2+(a+2b)^2+(2a+b)^2=(2a+2b)^2+a^2+b^2\\=5a^2+8ab+5b^2}$$
これで左辺と右辺の項数が$${3}$$個になる為,
理想的な解のパラメータ解となる.
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6.2 The ideal parameter solution in the case of k=3
【k=3の場合の理想的なパラメータ解】
$${\small \rm \bf 6.2~~The~ideal~parameter~solution~in~the}$$
$${\small \rm \bf case~of}$$ $${\bf k=3}$$
$${\small \rm To~the~identity~of~the~basic~solution~of}$$
$${\small \rm the~previous}$$ $${k=3}$$
$${\small \rm If~you~substitute}$$ $${d=a+b,c=a+2b}$$
$${\small \rm and~reduce~the~common~term~on~the~right}$$
$${\small \rm side~of~the~left~side,it~will~be~as~follows.}$$
$${0^k+(4a+3b)^k+(3a+b)^k+(a+2b)^k\\=(4a+2b)^k+(3a+3b)^k+a^k+b^k}$$
$${\small \rm Actually, I'll~try~to~calculate~by~assigging}$$.
$${k=1,2,3}$$.
$${k=1→}$$$${\small \rm (Left)=(Right)}$$$${=8a+6b}$$
$${k=2→}$$$${\small \rm (Left)=(Right)}$$
$${~~~=2(13a^2+17ab+7b^2)}$$
$${k=3→}$$$${\small \rm (Left)=(Right)}$$
$${~~~=(4a+3b)(23a^2+27ab+12b^2)}$$
$${\small \rm Because~this~makes~the~number~of~terms}$$
$${\small \rm on~the~left~and~right~sides}$$ $${4}$$,
$${\small \rm It~is~a~parameter~solution~of~the~ideal~solution.}$$
先の$${k=3}$$の基本的解の恒等式に
$${d=a+b,c=a+2b}$$を代入して左辺右辺の共通項を減らすと次の様になる.
$${0^k+(4a+3b)^k+(3a+b)^k+(a+2b)^k\\=(4a+2b)^k+(3a+3b)^k+a^k+b^k}$$
実際に$${k=1,2,3}$$を代入して計算してみる
$${k=1→}$$$${(}$$左辺$${)=(}$$右辺$${)=8a+6b}$$
$${k=2→}$$ $${(}$$左辺$${)=(}$$右辺$${)\\~~~=2(13a^2+17ab+7b^2)}$$
$${k=3→}$$ $${(}$$左辺$${)=(}$$右辺$${)\\~~~=(4a+3b)(23a^2+27ab+12b^2)}$$
これで左辺と右辺の項数が$${4}$$個になる為,
理想的な解のパラメータ解となる.
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$${REFERENCES}$$
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6.3 The ideal parameter solution in the case of k=4
【k=4の場合の理想的なパラメータ解】
The rest will be posted at a later date.
続きは後日に投稿予定です.
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