是否能用初中知识证明x^3+y^3=z^3无正整数解

发布网友发布时间：2023-12-27 16:20

共5个回答

热心网友时间：2024-07-15 01:04

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

热心网友时间：2024-07-15 01:04

热心网友时间：2024-07-15 01:05

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。。

热心网友时间：2024-07-15 01:05

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*，X与Y适用于任意整数，

热心网友时间：2024-07-15 01:06

应该不能吧

例：
X=3,Y=1,Z=4

3*3+1*3=4*3