在十六世纪早期,三次方程的解法是数学自希腊时代以来第一个毫不含糊的进步,它揭示了希腊人未曾驾驭的代数的威力,这种威力很快为几何开辟出一条新路,一条真正的王者之路。卡丹(Cardan)对这一发现的兴奋之情是完全可以理解的。即便在今天,任何一个人能亲自发现三次方程的解,至少对他高贵的数学生涯是一种激励。(参见卡茨(Mark Kac)(1984))
         ——《数学及其历史》   (by John Stillwell)

 

  

     在我们这个时代,博洛尼亚的费罗已经解决了三次幂和一次幂等于一个常数的情形,是非常巧妙和令人赞叹的成就。因为这门艺术的精妙与明晰超越了人类的一切能力,它真是来自天国的礼物,能够清楚地测定人的智力。任何人只要专注于它,就会相信世间没有任何事是不可理解的。

            ——卡丹(Cardan)

   

          
 
  费罗怎么会知道要这样做?美籍波兰数学家卡茨(Mark Kac, 1914-1984)用它关于普通天才与神奇天才之区别的名言回答了这个问题:“普通天才是你我差不多都可以成为的人,只要我们能再优秀上好几倍,关于他的心智的如何运作的,毫无神秘可言,一旦我们了解了他的所做,我们就会确信我们本来也是可以做到的,神奇天才则不同……他们心智的运作几乎完全不能理解,甚至在我们了解了他们的所做之后,他们对这些事的过程对我们来说仍完全是漆黑一片。”费罗的想法就属于神奇之类。

               ——《虚数之谜》

   


    人总要呆在一种什么东西中,沉溺其中,苟有所得,才能实证自己的存在,切实地掂出自己的价值。……人总要有点东西,活着才有意义。人总要把自己生命的精华都调动出来,倾力一搏,像干将、莫邪一样,把自己炼进自己的剑里,这,才叫活着。
                      ——汪曾祺

 

 

    想像力比知识更重要,因为知识是有限的,而想象力概括着世界上的一切,推动着进步。

       ——爱因斯坦(Einstein)

 

 

 

   数学是一种精神,一种理性的精神,正是这种精神,激发、促进、鼓舞并驱使人类的思维得以运用到最完美的程度。

                            ——克莱因(Klein

      

 

    没有一门其他的学科象数学一样有着如此清晰明确和被一致公认的标准,而那些被铭记的人几乎都是受之无愧的。数学上的荣誉,如果你有支付得起的现金的话,是最合理最稳固的投资之一。

           ——哈代G.H.Hardy

 

 

 

   能够作出数学发现的人,是具有感受数学中的秩序、和谐、整齐和神秘之美的能力的人。

                      ——庞加莱(Poincare

 

 

 

   在人类的心灵深处,都有一种根深蒂固的需要,这就是希望自己是个发现者、研究者和探求者。  

                              ——苏霍姆林斯基




   一个精巧的数学证明,精神上近乎一首诗。  

         ——克莱因(Klein

   到底是什么使我们感到一个解法、一个证明优美呢?那就是各个部分间的和谐、对称、恰到好处的平衡。  

                ——庞加莱(Poincare

   

   正如人类的每项事业都追求确定的目标一样,数学研究也需要自己的问题,正是通过这些问题的解决,研究者锻炼其钢铁般的意志和力量,发现新方法和新观点,达到更为广阔和自由的境界。     

       ——希尔伯特(Hilbert)

   科学的探讨和研究,其本身就含有至美,其本身给人的愉快就是酬报,所以我在我的工作里面寻得了快乐。  

          ——居里夫人

        (Madame Curie)

     人们之所以从事研究工作可以举出许多十分体面堂皇的动机,其中有三条远比其余的重要。 首先(舍此其余根本不能成立)是理性的好奇心,探求真理的强烈欲望。

                       ——哈代G.H.Hardy

      如果理智上的好奇,职业上的自豪,和抱负心是研究工作的主要激励物,那么可以肯定没有谁比数学家享有更好地满足这些欲望的机会了。他的研究对象是世上最激发好奇心的——没有哪一个领域里的真理爱跟人耍如此奇诡的把戏。

        ——哈代G.H.Hardy

    我们所作的或许微不足道,但它具有某种永恒的特征;而创造了具有一丝一毫永久性兴趣的任何东西,管它是几行诗句也好或者一个几何定理也罢,已经是做了某件完全超出芸芸众生绝大多数人能力之上的事情。

                           ——哈代G.H.Hardy


       
       
       

          菲尔兹奖得主Timothy Gowers论如何解三次方程                                        

                                                                             ——谢国芳(Roy Xie)译注                 Email:  roixie@163.com

                                                     如何自主地发现三次方程的解法

        让我们想像自己面对着一个三次方程 x3 + ax2 +bx +c = 0. 解出该方程意味着要写出一个求它的根的公式,该公式应该以它的系数a, b, c和一些常数 (即不依赖于a, b, c的数) 表示,并且只用加减乘除和开方运算。

        正如我在其他网页里所做的那样,我将表明这样的一个公式可以凭着标准的数学直觉推导出来,而不需要神秘的灵感闪现。我当然不是断言任何有理性的人都能在一两个小时内推导出这个公式——通常需要尝试几种不成功的直觉才能发现正确的标准化得数学直觉。然而,在任何给定的情况下,合适的直觉列表一般不会太长。如果你年轻,雄心勃勃,仍然不知道如何解三次方程,那么我建议你动手一试,或者在读一点本页的内容之后再作尝试,你在几个小时内获得成功的可能性很可能比你预想的高。

        让我们从一个数学中最普遍有效(而且明显易懂)的解题原则开始吧。

       如果你试图解决一个问题,看看能不能把一个已知的解法类推应用于一个类似的问题。

      运用这个原则可以避免对每一个新问题都从头开始。重要的并不是该问题本身的难度,而是克服该问题和其他已经解决的问题之间的差异难度。

       

                                             How to discover for yourself the solution of the cubic

    Let us imagine ourselves faced with a cubic equation x3 + ax2 +bx +c = 0. To solve this equation means to write down a formula for its roots, where the formula should be an expression built out of the coefficients a, b and c and fixed real numbers (that is, numbers that do not depend on a, b and c) using only addition, subtraction, multiplication, division and the extraction of roots.

    As I have done in other pages, I shall try to show that it is possible to derive such a formula by following standard mathematical instincts, without the need for mysterious flashes of inspiration. I am certainly not claiming that any sensible person should be able to derive the formula in an hour or two - finding the right `standard mathematical instinct' normally involves trying several that do not work. Nevertheless, the list of suitable ones to try in any given situation is usually not too long. If you are young and ambitious and do not yet know how to solve cubics, I would recommend having a go, or perhaps reading a short way into this page and then having a go. Your chances of succeeding in a few hours are probably higher than you think.


    Let us begin with one of the most useful (and obvious) general problem-solving principles in mathematics.

If you are trying to solve a problem, see if you can adapt a solution you know to a similar problem.

By using this principle, one can avoid starting from scratch with each new problem. What matters is not the difficulty of the problem itself but the difficulty of the difference between the problem and other problems whose solutions are known.  


                                                          二次方程的解法

        在现在这个情形中,显而易见,我们想到的类似的问题就是解二次方程x2 + 2ax + b = 0 (我加上因子2仅仅是为了方便,当然这在数学上没有任何区别)。我们怎么办呢?唔,我们"注意到"

                                                                 x2 + 2ax +b = (x+a)2 + b-a2

        这很快就导出解

                                                                  x = -a +/- (a2 -b)1/2

       这一招高明吗?在接下去考虑三次方程之前详细考察这个更初等的方程是有益的。所以让我们假想我们甚至不知道如何解二次方程。一个可能把我们引向它的解的思路是这样的:在干瞪着一般的方程x2 + 2ax +b = 0 毫无头绪之后,我们退回到下面这个问题:

       有我知道如何求解的特殊情形吗?

         然后,我们有点尴尬地注意到当a = 0时我们能解这个方程,也就是说,我们能解方程x2 + b = 0 (因为我们可以开平方根)。接下去,我们也许注意到如果b=a2 那么我们就得到了方程x2 + 2ax + a2 = 0, 它可以改写为(x+a)2 = 0. 一旦注意到这一点,我们就会认识到有帮助的并不是方程的右边是0,而是左边是一个完全平方,所以我们对于任意的b都能解出 (x+a)2=b ,这给了我们一大类能解出的二次方程,所以我们不问下面这个问题就太蠢了:

        有不能改写成 (x+a)2=b这种形式的二次方程吗?

       为了回答这个问题,我们需要把它重新写回原来的形式,这只要乘出括号,把b移到方程的左边就行了,这样我们就得到了方程x2 + 2ax + a2-b = 0. 到此就非常清楚了,我们可以令2a等于任何一个我们需要的数,在这样做了之后,接着我们又可以令 a2-b 等于任何另一个我们需要的数,于是二次方程就解出来了。

      如果你觉看出方程x2 + 2ax + a2 = 0解是一个过高的要求,那么还有另外一条路径:想知道1+21/2 是否是一个代数数并不需要太多的好奇心,注意到如果 x=1+21/2 那么 (x-1)2 = 2也不需要多高的才华,只要把这个例子加以推广,你很快就会认识到形如(x+a)2=b 的方程是可解的。

 

Solving quadratics

In this case, it is absolutely obvious that the similar problem we should take is that of finding a solution to the quadratic equation x2 + 2ax + b = 0. (I have put in the factor 2 just for convenience - of course it makes no difference mathematically.) How do we do that? Well, we `observe' that

x2 + 2ax +b = (x+a)2 + b-a2

which leads quickly to the solution

x = -a +/- (a2 -b)1/2

Was that observation clever? It will be useful to dwell on this more elementary question before continuing with the cubic. So let us imagine that we do not even know how to solve quadratics. One line of thought that might lead us to a solution is the following. After staring at the general equation x2 + 2ax +b = 0 and having no ideas, we fall back on the following question.

Are there special cases that I do know how to solve?

Then, with some embarrassment, we note to ourselves that we can solve the equation when a = 0. That is, we can solve the equation x2 + b = 0 (because we are allowed to take square roots). Next, we perhaps note that if b=a2 then we have the equation x2 + 2ax + a2 = 0, which can be rewritten (x+a)2 = 0. As soon as we have noticed this, we will realize that what helps is not that the right hand side is zero, but that the left hand side is a perfect square. We can therefore solve (x+a)2=b for any b. This gives us a whole family of quadratics that we can solve, so we would be mad not ask the next question.

Are there quadratic equations that cannot be written in the form (x+a)2=b?

To answer that, we need to put it back in the original form, by multiplying out the bracket and taking b over to the left hand side. This gives us the equation x2 + 2ax + a2-b = 0. It is then clear that we can make 2a any number we want, and that, having done so, we can make a2-b any other number we want. So the quadratic is solved.

If you think that it was asking too much to notice that the equation x2 + 2ax + a2 = 0 could be solved, then here is another route. It doesn't take much curiosity to wonder whether 1+21/2 is an algebraic number, or much talent to notice that if x=1+21/2 then (x-1)2 = 2. Generalizing this example leads quickly to the observation that equations of the form (x+a)2=b can be solved.


 三次方程的初步简化

        什么是配方这一操作在三次方程中的自然推广呢?要回答此类问题,下面这一策略常常是有用的:

        对你想要推广的东西给出一个一般性的描述。

      我将尝试直接通过实例来阐明我的意思。为了配方,我们注意到

                                  (x+a/2)2 = x2 + ax +a2/4

      因此我们可以把任何以 x2 + ax开始的二次方程写成 (x+a/2)2 加一个常数的形式。换一种说法是,如果我们令 y=x+a/2,那么 y就满足一个形式特别简单的二次方程 y2+C=0。当然,一旦我们解出了这个关于y的方程,很容易就得到x的解,因为x是y的一个很简单的一次函数。

      在这个关于y的这个方程什么变得更简单了呢?对于这个问题有两个合理的回答,把两者都考察一下是值得的。 第一个回答是注意到这个关于y的方程只包含y2 和一个常数项——所以用y代换x就使得我们可以设一次项的系数为0。第二个回答更加简明易懂——它更简单是因为我们断言形如y2+C=0的方程可以轻松求解。

    这一思路引发出两个问题。

 

A preliminary simplification of the cubic.

What would be the natural generalization to cubics of the process of completing the square? To answer a question of this kind, the following tactic is often useful.

Give a general description of what it is that one would like to generalize.

I shall try to illustrate what I mean just by doing it. To complete the square one notes that (x+a/2)2 = x2 + ax +a2/4, so that we can write any quadratic that begins x2 + ax as (x+a/2)2 plus a constant. To put that another way, if we let y=x+a/2, then y satisfies a quadratic equation of the particularly simple form y2+C=0. Of course, once we have solved the equation for y, it is easy to obtain a solution for x, since x is a very simple linear function of y.

What was simpler about the equation for y? There are two reasonable answers to this question, and it is worth looking at both of them. The first is to note that the equation for y involves only y2 and a constant term - so replacing x by y allows us to assume that the coefficient of the linear term is zero. The second is more obvious - it is simpler because by allowing ourselves to take square roots we have declared that equations of the form y2+C=0 can be solved at a stroke.

This line of thought leads to two questions.

 

 

1. Is there a similar way to simplify a cubic so that some of the coefficients become zero?

2. Is there a similar way to simplify a cubic so that it becomes of the form y3+C=0?

The answer to question 1 is not hard to find. If y=x+t then y3=x3+3tx2 +3t2x+t3. Therefore, if t=a/3 then the cubic x3 + ax2 +bx +c can be rewritten as y3 + py +q, where (for what it is worth) p=b-3t2 and q=c-bt+2t3. Writing this in terms of a we have p=b-a2/3 and q=c-ab/3+2a3/27.

As for the second question, we can start to think about it by asking ourselves the following direct generalization of a question we asked about quadratics.

Are there cubic equations that cannot be written in the form (x+a)3=b, and if so, which ones can?

Expanding and subtracting, we find that we can easily solve equations of the form

x3 + 3ax2 + 3a2x + a3 - b = 0

When is the equation

x3 + ax2 + bx +c = 0

of this type? Comparing it with the preceding one we see that it is of the required form if the pair (a,b) is of the form (3s,3s2) for some s, which it is if and only if a2=3b. So the following question arises naturally.

Can we replace x by some y=x+t in such a way that y satisfies a cubic with a'2=3b' (where a' and b' are the coefficients of y2 and y respectively).

    1.有类似的方法可以简化一个三次方程使得它的某些项的系数变成0吗?

    2.有类似的方法可以简化一个三次方程使得它变成 y3+C=0 的形式吗?

    第一个问题的答案是不难找到的。如果y=x+t那么y3=x3+3tx2 +3t2x+t3. 因此,如果t=a/3 ,那么三次方程x3 + ax2 +bx +c 可以改写为 y3 + py +q,其中p=b-3t2q=c-bt+2t3,用a表示就有p=b-a2/3 q=c-ab/3+2a3/27

    至于第二个问题,我们的最初想法是问我们自己下面这个问题,它是我们前面对二次方程所问问题的直接推广:

    有不能改写成 (x+a)3=b这种形式的三次方程吗?如果有,那么那些三次方程可以改写成这种形式呢?

    将(x+a)3=b展开,把b移到方程左边,我们发现我们能很容易地解出下面这一形式的三次方程:

x3 + 3ax2 + 3a2x + a3 - b = 0

    什么时候方程

x3 + ax2 + bx +c = 0

属于这一类型呢? 比较两者我们发现它将具有我们所要求的形式,如果数组 (a,b)形如 (3s,3s2),其中s是某个数,显然当且仅当a2=3b.时才会如此,于是下面这个问题就自然产生了:

    我们能用某个y=x+t 替换x使得y满足a'2=3b' 的三次方程吗?(a' b' 分别是y2 y 的系数) 

Comparing it with the preceding one we see that it is of the required form if the pair (a,b) is of the  form (3s,3s2) for some s, which it is if and only if a2=3b. So the following question arises naturally.

Can we replace x by some y=x+t in such a way that y satisfies a cubic with a'2=3b' (where a' and b' are the coefficients of y2 and y respectively).

 

 

 

    这一思路看上去是很有希望的,因为t给了我们一个自由度,而我们所需要的只不过是一个条件——即a'2-3b' 等于0,回答这个问题的方法是显而易见的,所以就让我们动手尝试吧。作代换 x=y-t我们就得到了方程

(y-t)3 + a(y-t)2 + b(y-t) +c = 0

重新整理后它可以改写成

y3 + (a-3t)y2 + (b-2at+3t2y) + c-bt+at2+t3

它告诉我们a'=a-3t b'=b-2at+3t2.,因此

a'2-3b'=a2-6at+9t2 -3b+6at-9t2=a2-3b.

 我们所表明的是我们不可能通过形如 y=x+t.的代换改变量a2-3b ,换句话说,对上面问题2的回答是否定的,至少”类似的方法“被理解成用这样一个代换的话,量a2-3b 不变的一个稍微花俏的说法是把它称为一个不变量。

    a2-3b 是一个不变量是一个不幸的偶然事件吗?更深一层的思考将为我们揭示产生这一现象的原因,并表明如果我们期望能这样简单地解出三次方程那就太愚蠢了。你也许已经注意到a2-3b=-3p, 其中p是当我们把三次方程x3 + ax2 + bx +c变成更简单的三次方程 y3 + py + q. 后一次项的系数,我们选择y等于 x+a/3,易见没有任何其他的选择能使得y2 的系数变成0。因此,我们发现的不变量有一个解释(正如你应该一直预料/期望的ddd),它是当二次项通过一个形如 y=x+t的代换被消去后一次项系数。

   但现在很显然这个量是不变量,归根结底。。。如果我作代换y=x+s(s为任意数),然后问什么样的进一步的代换z=y+r 能消去二次项,回答z=x+r+sr+s必须等于 a/3,因此,对于y所得到的p和对于x所得到的p是一样的

 

This approach looks promising, because it gives us one degree of freedom and all we want is one condition - that a'2-3b' should be zero. It is obvious how to answer the question, so let us go ahead and do it. Writing x=y-t and substituting we obtain the equation

(y-t)3 + a(y-t)2 + b(y-t) +c = 0

which rearranges itself to

y3 + (a-3t)y2 + (b-2at+3t2y) + c-bt+at2+t3

This gives us a'=a-3t and b'=b-2at+3t2. Therefore,

a'2-3b'=a2-6at+9t2 -3b+6at-9t2=a2-3b.

What we have shown is that we cannot change the quantity a2-3b by making a substitution of the form y=x+t. In other words, the answer to question 2 above is no, at least when `a similar way' is taken to mean that we should use such a substitution. A slightly fancier way to say that a2-3b doesn't change is to call it an invariant .

Is it an unfortunate accident that a2-3b is an invariant? Further reflection gives us a reason for this phenomenon, and shows that we were foolish ever to expect that the cubic could be solved so simply. You may already have noticed that a2-3b=-3p, where p was the coefficient of the linear term we obtained when we converted the cubic x3 + ax2 + bx +c into the simpler cubic y3 + py + q. We chose y to be x+a/3 and it is easy to see that no other choice would have led to the coefficient of y2 being zero. Hence, the invariant we have discovered has an interpretation (as one should always expect): it is the coefficient of the linear term when the quadratic term has been removed by a substitution of the form y=x+t.

But it is now obvious that this quantity is an invariant. After all, if I substitute y=x+s (for any s) and then ask what further substitution z=y+r will remove the quadratic term, the answer is that z=x+r+s and r+s has to be a/3. Therefore, the p that I obtain for y is the same as the p that I obtain for x.


    一个僵局和如何打破这个僵局 

       实际上,一开始就很明显,上面第二种解三次方程的办法注定是要行不通的。因为假如“配立方”是可能的,那么所有三次方程都将是 (x+a)3+b的形式,但如果是这样,那么为什么我们还要费力劳神地把一个三次方程转换成这种形式呢?所以,“配立方”不只是不可能的,而且它是基于简单而无可争辩的原因而不可能的。另一方面,“配立方”难道不是“配平方”的自然推广吗?现在既然我们在尝试之后失败了,似乎我们解三次方程的最大机会(那就是考察我们如何解二次方程然后进行类推)也告吹了

      然而,这种失败主义的态度常常是错误的,也许我们可以用另一个普遍原则来表达这种观点:

      可能有很多种方法进行类推或推广一个证明。

      但是你会问,我们应该怎么样搜索不同的推广呢?让我稍微修改一下前面我给出的一个建议。 

    给出你想要推广的论断的一个描述。解释它为什么成立。把解释变得更含糊、更一般,然后试图寻找基于同样的(更含糊的)理由也成立的不同的论断。

    为了能把它付诸实行,让我再一次描述如何解二次方程。

 

 

An impasse and how to break it.

Actually, it was obvious in advance that the second approach to solving the cubic was doomed to fail, since if it were possible to `complete the cube' then every cubic would be of the form (x+a)3+b. But if that were true, then why would we have been bothering to convert a cubic into that form? So, not only is completing the cube impossible, it is impossible for simple and compelling reasons. On the other hand, isn't completing the cube the natural generalization of completing the square? Now that we have tried it and failed, it is as though we have blown our main chance to solve the cubic (which was to see how we solved the quadratic and adapt our method).

However, this sort of defeatist attitude is often a mistake. Perhaps one can even express this view with another general principle.

There may be many ways to adapt or generalize a proof.

But how, one might ask, should one search for different generalizations? Let me modify an earlier suggestion.

Give a description of the argument that one would like to generalize. Explain why it worked. Make the explanation vaguer and more general, and then try to find different arguments that work for the same (vaguer) reasons.

So that we can put that into practice, let me once again describe how to solve the quadratic.

 

 

 

y=x+a/2,那么 y就满足一个形式特别简单的二次方程 y2+C=0一旦我们解出了这个关于y的方程,很容易就得到原方程x的解,因为x是y的一个很简单的一次函数。

我们需要y的两个性质。第一,y应该满足一个我们知道如何求解的方程,第二,x应该以一种简单的方式由y决定——使得我们一旦知道了y就能算出x。

 

 

 

Let y=x+a/2. Then y satisfies a quadratic equation of the particularly simple form y2+C=0. Once we have solved this equation for y, it is easy to obtain a solution of the original equation for x, since x is a very simple linear function of y.

Why, in general terms, did that work? We needed two properties of y. First, y should satisfy an equation that we knew how to solve, and secondly x should depend on y in a simple way - so that once we knew y we could work out x.

If we wish to carry this approach over to the cubic, then we should have clear answers to the following two questions.

(i) Which equations can we solve?

(ii) How are we prepared to allow y to depend on x?

The answer to the first question we more or less know already. We can solve linear and quadratic equations, and also cubic equations if they happen to have the nice form x3+C=0. As for the second, so far we have considered substitutions of the form y=x+t. What other substitutions could there possibly be?

I shall answer this question by yet another time-honoured method, which occurs all over mathematics.

Do the most general thing you can possibly imagine. Then, when you find that you need certain properties, make what you have done more specific by introducing those properties.

Suppose then that we make the substitution y=f(x). (It is hard to see how we could be more general than that.) Let us suppose that this leads to an equation for y that we can solve. When will knowing y be of any use? The answer is obvious - when we can solve the equation y=f(x) for x in terms of y. But we know which equations we can solve - linear, quadratic and simple cubic equations. We have already tried linear substitutions and seen their limitations, so we are left with two reasonable possibilities for f(x). One is x2+ax+b (it is not hard to see that giving x2 a different coefficient is not going to make a significant difference) and the other is x3+c.

Following some very general problem-solving techniques has led us to an idea that is definitely new. By standing back a little, we realized that the important thing about the substitution y=x+t in the solution of the quadratic was not that it magically worked, or that it was linear, but that it was invertible in the sense that we could give a formula for x in terms of y. The impasse is now broken in the sense that we have an approach to try with the cubic. It may not work, but having an approach that may or may not work is much better than having no approach at all.


A substitution that solves the cubic.

If a linear substitution worked for quadratic equations, then which sounds more likely to work for cubic equations - a quadratic substitution or a particular kind of cubic substitution? Somehow the quadratic one is more promising, as it fits the general description of having degree one less than that of the equation one is trying to solve. This is not a particularly convincing argument, but the worst that can happen is that we try it and it doesn't work. So let us see where we can get with the substitution y=x2+ux+v.

如何一个线性代换对二次方程奏效,那么对于三次方程哪一个看上去更有可能奏效——是一个二次代换还是一个特别的三次代换呢?不知为什么,直觉上/二次代换是更有希望的,因为它符合一般的描述,即比我们试图解的方程少一个自由度,这并不是一个特别令人信服的论据,但可能发生的最坏情况是我们尝试之后失败了。所以让我们看看利用代换 y=x2+ux+v我们能得到什么结果。

我们现在碰到了一个问题,我们希望y将满足一个形式特别简单的三次方程,但它满足任何一个三次方程是显然的吗?如果你不觉得这是显然的,那么现在。。。。。。,因为在那里我反复使用了一个技巧,它在这里也同样奏效。。。

我们知道x满足方程

x3 + ax2 + bx +c = 0

这意味着每次我们写下一个x的多项式,我们可以用 -ax2-bx-c代替x3 ,  -ax3-bx2-cx 代替x4 等等,这就是说,任何一个x的多项式都等于一个x的二次函数。但y2 y3是x的多项式函数,因此就等于x的二次式,这对于1和y也平凡地成立,因此数 1, y, y2 y3 都具有形式 rx2+sx+t. 欲使y满足一个三次方程,我们需要一个 1, y, y2 y3 的一个非平凡的////的线组合等于0,为此我们需要解三个关于四个未知数的齐次线性方程,这总是能办到的。

We now run into a problem. We are hoping that y will satisfy a cubic of a particularly simple form. But is it obvious that it satisfies any cubic? If you do not find it obvious then this is the point where it will help to have read my page on algebraic numbers , because there I repeatedly used a trick which works here as well (and which, I stress, arose naturally in that context).

We know that x satisfies the equation

x3 + ax2 + bx +c = 0

But this means that every time we write down a polynomial in x, we can replace x3 by -ax2-bx-c, x4 by -ax3-bx2-cx and so on. That is, every polynomial expression in x is equal to some quadratic function of x. But y2 and y3 are polynomial functions of x, and hence equal to quadratic ones. This is trivially true of 1 and y as well. Hence, the numbers 1, y, y2 and y3 are all of the form rx2+sx+t. For y to satisfy a cubic we need a non-trivial linear combination of 1, y, y2 and y3 to be zero. To obtain this, we need to solve three homogeneous linear equations in four unknowns, which we can always do.

 

 

 

So now we can describe a possible method more precisely: let y=x2+ux+v, work out y2 and y3 in terms of x, reduce them to quadratics using the fact that x3=-ax2-bx-c, find a non-trivial linear relationship between 1, y, y2 and y3, write out the corresponding cubic y3+dy2+ey+f in y and finally (the most important part) cleverly choose u and v in such a way that d2=3e.

现在我们能更准确地描述一个可能的方法了,令 y=x2+ux+v, 用x表示 y2 y3 ,利用x3=-ax2-bx-c,将它们归化为x的二次式,再找到一个 1, y, y2 y3 之间的非平凡的///线性关系,写出相应的三次方程y3+dy2+ey+f ,最后(最重要的一步)巧妙地选择u v 使得 d2=3e.

 

We have no guarantee that this will work, because it may be that, as happened with linear substitutions, there simply is no choice of u and v that makes d2 equal to 3e, or it may be that, although such a choice exists, the dependence of u and v on a, b and c is so complicated that we don't know how to solve the resulting equations. It is reasonable not to worry too much about the first potential difficulty, because we now have an extra degree of freedom, and there doesn't seem to be an argument telling us that it cannot possibly help. However, if you now go away and try to work out the details of the argument outlined above, you will see that complication is something one should definitely worry about. Indeed, it may seem after a while that in order to work out u and v you will have to solve quintics .

我们并没有任何保证这会成功,因为有可能,正如前面的线性代换的情形一样,

 

 

Let me take it as read that just plunging in is a bad idea. In any case, it is another good problem-solving strategy to try simpler (but less general) approaches first, just in case they work. So how might we make the above calculations manageable?

One obvious idea is to use the simplification we obtained earlier: we might as well assume that a=0. This will allow us to replace x3 by -px-q. In fact, it is a little nicer to say that x3=px+q, which we can do by changing the definitions of p and q. And what about the substitution y=x2+ux+v? Well, remembering the invariant we discovered earlier, we should realize that y satisfies a cubic that we can easily solve if and only if y-v does. So we might as well save on algebra by setting v=0. In other words, not only will we simplify calculations by setting y=x2+ux, we will not even lose any generality.

Here are some calculations that arise when one starts with the equation x3=px+q, sets y=x2+ux, and tries to find a cubic satisfied by y. Doing it directly (that is, by working out y2 and y3 and solving some simultaneous equations) still gets unpleasant, but the calculations can be kept manageable by simplifying as we go along, as is done below. I shall also save time by writing C to mean a constant (depending on p,q and u) which may vary from line to line.

x3=px+q

y=x2+ux

y2=x4+2ux3 +u2x2

=(u2+p)x2+(2up+q)x+2uq

=(u2+p)y+(up+q-u3)x+2uq

y3=(u2+p)y2 +(up+q-u3)xy+2uqy

=(u2+p)y2 +(up+q-u3)(x3+ux2)+2uqy

=(u2+p)y2 +(up+q-u3)(ux2+px)+2uqy+C

=(u2+p)y2 +(up+q-u3)(uy+(p-u2)x)+2uqy+C

From an earlier line we have that

(up+q-u3)x =y2-(u2+p)y+C

so this is equal to

(u2+p)y2 +(up+q-u3)uy+(p-u2)(y2-(u2+p)y) +2uqy+C

=2py2+(u2p+3uq-p2)y+C

Thus, y satisfies the cubic equation

y3-2py2 -(u2p+3uq-p2)y-C=0

and all that is left to decide is whether u can be chosen in such a way that

(-2p)2=-3(u2p+3uq-p2)

that is, such that

3pu2+9qu+p2=0

This, being a quadratic in u, can be solved. Using this value of u one obtains a cubic in y that can be `completed'. This gives a solution of y. Then x can be worked out from y by solving a further quadratic.

Of course, the resulting formula, if one worked it out, would be pretty unpleasant, and I should now say that better methods have been discovered (involving different substitutions) that lead to easier calculations and neater answers. They are easy to find on the internet, but all tend to have a `magic' quality about them. I should also say that I haven't discussed the annoying problem that not all the `solutions' that arise by the above method will necessarily be solutions, since knowledge of y doesn't uniquely determine x.

Just to see how there might be other sensible substitutions, let us return to the question of which ones allow us to calculate x. We noted that we could do so if y was a quadratic function of x. But this was not absolutely necessary, even if we could only solve quadratic equations. For example, if y were defined implicitly by x2+uxy+v=0, then knowing y would still allow us to determine x.

As a matter of fact, the simplest method works for a completely different reason. It involves substituting the other way - by setting x=w+p/3w (when the equation is x3=px+q) and finding that the resulting equation in w is a quadratic in w3. This method is described in more detail here . I do not have a plausible explanation for how it was discovered.


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