是否能用初中知识证明x^3+y^3=z^3无正整数解

2024-12-29 17:34:40

推荐回答（5个）

回答1：

Solving x^3 + y^3 + z^3
Date: 02/22/2003 at 08:06:53
From: Martin
Subject: Solving x^3 + y^3 + z^3

Given: x + y + z = 0; x * y * z = 2; x,y,z are not equal to each
other; x,y,z all real numbers.

Solve for x^3 + y^3 + z^3

When doing the algebra, what to do with the terms with x^2, y^2, or
z^2 ?

I start off with (x + y + z) = 0; therefore (x + y + z)^3 = 0.
Multiplying out yields x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3y^2x +
3y^2z + 3z^2x + 3z^2y + 6x*y*z. I keep the x^3, y^3, and z^3 on the
same side of the equals sign. I can then solve for their sum. On the
other side, with ease I substitute 2 into the 6x*y*z = 6*2.

The problem is what to do with the terms such as x^2y or y^2x, etc.
I have tried every way of factoring hoping to be left with something
I can work with. I also tried substituting in from (x + y + z) = 0 and
x * y * z = 2. The factoring and substitution did not yield anything I
can use. Is there a trick to dealing with the terms such as y^2z or
did I take the wrong approach from the beginning?

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Date: 02/22/2003 at 15:29:18
From: Doctor Greenie
Subject: Re: Solving x^3 + y^3 + z^3

Hi, Martin -

(1) Your approach to the whole problem is exactly on target.
(2) Yes, there IS a trick - a really cool trick! - for dealing with
the terms such as x^2y.

We have

0 = (x+y+z)^3 = (x^3+y^3+z^3)+3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2)+6xyz

and so

x^3+y^3+z^3 = -3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - 6xyz
= -3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - 12

Now our task is to evaluate the expression

x^2y+x^2z+xy^2+y^2z+xz^2+yz^2

To do this, let's start by grouping terms and factoring so that there
are no "squared" terms left:

x^2y+x^2z+xy^2+y^2z+xz^2+yz^2
= x(xy+xz) + y(xy+yz) + z(xz+yz)

And here comes the trick....

In this factored form, if we could get the three expressions in
parentheses to be identical, then we would have

x(...) + y(...) + z(...) = (x+y+z)(...) = 0(...) = 0

So let's add some terms (and then subtract those same terms) to do
exactly that:

x^2y+x^2z+xy^2+y^2z+xz^2+yz^2
= x(xy+xz) + y(xy+yz) + z(xz+yz)
= x(xy+xz+yz-yz) + y(xy+yz+xz-xz) + z(xz+yz+xy-xy)
= x(xy+xz+yz) + y(xy+xz+yz) + z(xy+xz+yz) - x(yz) - y(xz) - z(xy)
= (x+y+z)(xy+xz+yz) - 3xyz
= 0(xy+xz+yz) - 3(2)
= 0 - 6
= -6

And then we are done:

x^3+y^3+z^3 = -3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - 6xyz
= -3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - 12
= -3(-6) - 12
= 18 - 12
= 6

I hope all this helps. Please write back if you have any further
questions about any of this.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/

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Date: 02/22/2003 at 16:35:53
From: Martin
Subject: Thank you (solving x^3 + y^3+z^3)

Dr. Greenie,

In the trick in the factoring that you demonstrated to me and I
missed, it is evident what a beautiful subject math is!

Thank you very much for your trouble. Both you and the whole Dr. Math
group should NEVER be taken for granted.
Best wishes,
Martin

回答2：

回答3：

证明不了。这个是费尔马大定理。不光初中、高中无法证明，就连一般大学生也不会。这个要用到很多高等数学思想的。。

回答4：

能既然这个等式成立，那么它适用于任意整数，就像X+Y=3 适用于任意整数，令X为任一整数如1，同样可令Y为2，再将x^3+y^3 算出来，再开立方，是整数，就成立，这个题就是要认识到定义域无限制，X与Y适用于任意整数，

回答5：

应该不能吧

例：
X=3,Y=1,Z=4

3*3+1*3=4*3