# [转载] 希尔伯特第十问题: 一段数学发现史

1900 年夏天，第二届国际数学大会在巴黎举行，全球数学家欢聚一堂。大卫 · 希尔伯特（ 1862 1943 ），著名的德国数学家，哥廷根大学教授，应邀在大会上作主要的演讲。 作为世界上最伟大的数学家之一，他以在代数理论和数论研究工作上的建树而闻名，特别是大会前不久在基础性研究《基础几何》中重新建立欧几里德几何的公理系统，更使他再次名声斐然。 经过长期的思考，他选择一种别开生面的方式来开展他的演讲。在演讲中，他使用 数学问题 来定义那些在他认为会决定下个世纪数学发展的数学问题。 In the summer of 1900 mathematicians met on the Second International Congress in Paris. David Hilbert (1862-1943), the famous German mathematician, Professor of the Goettingen University, was invited to deliver one of the main lectures. As the greatest World mathematician he became famous by his works in algebra and number theory, and shortly before the Congress resolutely, he has rebuilt an axiomatics of the Euclidean geometry in the fundamental work "Foundations of Geometry" (1899). After long doubts Hilbert chose an unusual form of the lecture. In the speech "Mathematical Problems" he decided to formulate those mathematical problems, which, in his opinion, should determine development of mathematics in the upcoming century.

1900 年国际数学大会中，希尔伯特的演讲也许是数学家们听到的演讲中最有影响力的，也是数学家的演讲中最有影响力的，更是传播数学的演讲中最有影响力的。希尔伯特提出了在下个世纪将被研究的 23 个主要的数学问题。有些涉及面广，如物理学公理化（问题 6 ），可能永远也不能完全解决。其它的，如问题 3 ，更具体和容易解决。有一些问题则是用与他预期截然相反的方法来解决的，如 连续统假设 （问题 1 ）。 Hilbert's address of 1900 to the International Congress of Mathematicians in Paris is perhaps the most influential speech ever given to mathematicians, given by a mathematician, or given about mathematics. In it, Hilbert outlined 23 major mathematical problems to be studied in the coming century. Some are broad, such as the axiomatization of physics (problem 6) and might never be considered completed. Others, such as problem 3, were much more specific and solved quickly. Some were resolved contrary to Hilbert's expectations, as the continuum hypothesis (problem 1). [  译者注 ]  ：倍立方问题：求一立方体的棱长，使其体积是已知立方体的二倍；三等分角问题：求一角，使其角度是已知角的三分之一；化圆为方问题：求一正方形的边长，使其面积与一已知圆的相等。

x = (17y + 1)/7 = 2y + (3y + 1)/7.

(3y+1)/7 必须是整数，标记为 z 。于是有 3y + 1 = 7z  和  x = 2y + z 。从而我们推导出来的方程 3y - 7y = -1 的系数比最初的小。再次使用 降系数法 得： (The number (3y + 1)/7 must be integer, let us denote it by z. Then 3y + 1 = 7z and x = 2y + z. Thus we have arrived at the equation 3y - 7z = -1 having smaller coefficients than the initial one. Let us apply our "coefficient reduction method" once more:)

y = (7z - 1)/3 = 2z + (z - 1)/3.

(z-1)/3 必须是整数，标记为 t 。于是有 z=3t+1 ，(并且 The number (z - 1)/3 must be integer, let us denote it by t. Then z = 3t + 1, and )

y = 2z + t = 7t + 2,

x = 2y + z = 2(7t + 2) + 3t + 1 = 17t + 5.

x = 5, y = 2; x = 22, y = 9; x = -12, y = -5; x = 39, y = 16; ... .

[ 译者注 ] ：费马最后定理在中国又称为 费马大定理

x = m 2 - n 2 , y = 2mn , z = m 2 + n 2 ,

x 3 + y 3 = z 3 , x 4 + y 4 = z 4 , x 5 + y 5 = z 5 , ... .

x n + y n = z n

n>2 时， x y z 没有非零整数解。换言之，方程在 n > 2 的情况下没有自然数解。 has no non-zero integer solutions for x, y and z when n > 2. In other words, this equation at n > 2 has no solutions in natural numbers. 1994 9 19 ，长达 350 年之久的费马猜想最终由英国数学家 安德鲁 · 怀尔斯证明了。 Fermat's 350 years old hypothesis was finally proved on September 19, 1994 by English mathematician Andrew Wiles.

1900 8 6 12 日，第二次国际数学大会在巴黎举行。在他八月 8 号即星期三早上的演讲中，大卫 . 希尔伯特阐述了他著名的 23 个数学问题（全文在http://aleph0.clarku.edu/~djoyce/hilbert/problems.html）。 23 个问题中的第十问题如下：

10.   判定丢番图方程的可解性

(From August 6th to August 12th 1900, the Second International Congress of Mathematicians took place in Paris. In his Wednesday morning lecture of August 8 David Hilbert stated his famous 23 mathematical problems for the coming XX century (see full text at http://aleph0.clarku.edu/~djoyce/hilbert/problems.html). The 10th of these 23 Hilbert's problems was the following:

10. Determining the solvability of a Diophantus equation.

Given a Diophantus equation with any number of unknowns and with rational integer coefficients: devise a process, which could determine by a finite number of operations whether the equation is solvable in rational integers.)

1936 年，图灵，波斯特和丘奇提出了第一个关于算法的形式化概念。显而易见，同时他们发现首个不可解的大量问题。此后不久即时 1950 年，马丁 ·  戴维斯   在他的博士论文（参阅http://cs.nyu.edu/cs/faculty/davism/）中向证明希尔伯特第十问题具有否定答案，即丢番图方程的不可解迈出了第一步。 In 1936 Turing, Post and Church, introduced the first formalized concepts of algorithm. Obviously, they also discovered the first unsolvable mass problems. Soon after this, in his Ph.D. theses of 1950 Martin Davis (see http://cs.nyu.edu/cs/faculty/davism/) made the first step to prove that Hilbert's Tenth problem is unsolvable. 1970 年马蒂亚塞维奇完成证明希尔伯特第十问题具有否定答案中最后缺失的一环后，他的名字在全世界变得广为人知。 The name of Yuri Matiyasevich became known worldwide in 1970 when he completed the last missing step in the "negative solution" of Hilbert's tenth problem. 想法如下：一般计算科学表示信息的工具使用单词而非数字。然而，使用数字来表示单词的方法有多种。其中有一种很自然地与丢番图方程关联。即不难证明任何   2 X 2 矩阵  "The idea was as follows. A universal computer science tool for representing information uses words rather than numbers. However, there are many ways to represent words by numbers. One of such methods is naturally related to Diophantus equations. Namely, it is not difficult to show that every 2 X 2 matrix   m 11 x m 22 - m 21 x m 12 = 1.

[ 译者注 ] 任何自然数均表示为任意不同的不连续的斐波纳契数之和，例如 30 可以表示为 30=21 + 8 + 1=21 x 11 +13 x 10 +8 x 11 +5 x 10 +3 x 10 +2 x 10 +1 x 11 。因此数字 30 对应的单词是 “1010001” 。由于表达中不存在连续的斐波纳契数，故对应的单中不存在连续的两个 “1”

“1969 年秋天，一位同事跟我说： 快去图书馆，朱莉 - 逻宾逊在最近一期的美国数学社会学报上发表了新的论文！ 但我早已把希尔伯特第十问题搁置一边了，对自己说 朱莉 - 逻宾逊在此问题上取得新的进展，这很好，但我不能再花时间在此了 ，故我没有去图书馆。  "One day in the autumn of 1969 some of my colleagues told me: "Rush to the library. In the recent issue of the Proceedings of the American Mathematical Society there is a new paper by Julia Robinson!" But I was firm in putting Hilbert's tenth problem aside. I told myself: "It's nice that Julia Robinson goes on with the problem, but I cannot waste my time on it any longer". So I did not rush to the library.

x 2 -   (a 2 -1) y 2 = 1.                                 (1)

c n+1 = 2a     c n     -     c n-1 ,                    (2)

f n+1 = 2a     f n   -     f n-1                       (3)

f 0 , f 1 , ..., f n , ...     (mod    a -1)               (4)

0,     1,      2,     ...,      a-2,

c0 - (a -2)  f 0 ,    c1 - (a - 2)  f1 ,     ...,  cn - (a -2) fn ,     (mod    4a-5)        (5)

20 , 21 , 22 , ... .

…… 读完朱莉 - 逻宾逊的论文后，我马上明白沃罗比约夫定理非常有用。 1970 年朱莉 - 逻宾逊从我这收到此书的副本，此前她一直没有读过沃罗比约夫那本书的第三版。如果沃罗比约夫把他的定理收录在此书的第一版里，谁能知道有什么事情发生呢？也许希尔伯特第十问题就提前十年解决了！ ” ... After I read Julia Robinson's paper I immediately saw that Vorob'ev's theorem could be very useful. Julia Robinson did not see the third edition of Vorob'v's book until she received a copy from me in 1970. Who can tell what would have happened if Vorob'ev had included his theorem in the first edition of his book? Perhaps Hilbert's tenth problem would have been "unsolved" a decade earlier!" References:

1944埃米爾·萊昂·珀斯特（英语：Emil Leon Post）首先猜測，對於第十問題，應尋求不可解的證明。
1949马丁·戴维斯利用库尔特·哥德尔的方法，並應用中國餘數定理的編碼技巧，得到遞歸可枚舉集的戴維斯範式
• {a|∃y∀k≤y∃x1,…,xn[p(a,k,y,x1,…,xn)=0]}{\displaystyle \{a|\exists y\,\forall k\!_{\leq y}\,\exists x_{1},\ldots ,x_{n}[p(a,k,y,x_{1},\ldots ,x_{n})=0]\}} 1950朱莉娅·罗宾逊在未知戴维斯工作的情況下，試圖證明冪函數是丟番圖的
• z=yx{\displaystyle z=y^{x}} • (a,b)∈D⇒b<aa{\displaystyle (a,b)\in D\Rightarrow b<a^{a}} • ∀k>0∃(a,b)∈D{\displaystyle \forall k>0\,\exists (a,b)\in D} 使得 1959戴维斯與普特南共同研究了指數丟番圖集，也就是以丟番圖方程所定義的集合，但其中指數可以是未知數。使用戴維斯範式與罗宾逊的方法，並且利用數論中一個當時尚未證明的假設（註：已於2004年由班·格林（英语：Ben Green）和陶哲軒所證明）：存在任意有限長度全由質數所組成的算數級數，他們證明了每一個遞歸可枚舉集都是指數丟番圖的。因此若是J.R.成立，就可以將「指數」兩字拿掉，而得到每一個遞歸可枚舉集都是丟番圖的。因而第十問題是不可解的。
1960罗宾逊證明了上述的數論假設是不必要的，並且大大簡化了證明。從而可知，只要能證明冪函數是丟番圖的，第十問題就可以解決。而關鍵又是尋找滿足J.R.假設的丟番圖集。
1961-1969戴维斯與普特南提出數種可證明J.R.的假定。罗宾逊指出，若存在一個全由質數組成的無限丟番圖集，便可證明J.R.
1970尤里·马季亚谢维奇指出可由十個一次和二次的聯立不定方程組，定義偶角標的斐波那契函數
• b=F2a{\displaystyle b=F_{2a}} http://blog.sciencenet.cn/blog-498408-1202676.html

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