My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Thanks Tree3Thanks
Reply
 
LinkBack Thread Tools Display Modes
July 1st, 2017, 08:30 AM   #11
Global Moderator
 
Joined: Dec 2006

Posts: 17,533
Thanks: 1322

It was 2³5 (which is 40), not 2*35.
skipjack is offline  
 
July 1st, 2017, 05:45 PM   #12
Senior Member
 
Joined: May 2016
From: USA

Posts: 691
Thanks: 284

Quote:
Originally Posted by uvkajed View Post
...if you think you can:

Theorem: x^n + y^n = z^n has no solution in positive
integers x, y, z if n > 1 is any odd integer.

By contradiction, suppose some solution in Integers
xyz > 0 and n > 1 odd.

The proofs of Claims 1) and 2) below
are given at the end of the argument.

1) x + y > z if n > 1.
2) 2z > x + y if n > 0.
3) x + y is a factor of x^n + y^n
if and only if n is an odd integer.

By 3), x^n + y^n = (x + y)[Q(x, y)] = z^n, where Q(x, y)
is some polynomial - the sum of terms of which by
hypothesis must necessarily be some integer s > 1.
Hence we can write (x + y)s = z^n. It follows from 1)
that x + y can not be a factor of z, and it follows from
2) that x + y can not be some power of z; therefore
x + y can not divide z^n, and it follows that s can not
be an integer as was supposed. Contradiction.
***

1) Substitute x + y for z and expand:
x^n + y^n = (x + y)^n = x^n + P(x, y) + py^n, where
the sum of terms of the polynomial P(x, y) must
necessarily be some positive integer r. From the
preceding, 0 = r. Contradiction. It follows that z
can not equal (or be greater than) x + y. Hence
z < x + y.

2) By symmetry we can suppose x < y. Then
y < z and 2y < 2z implies x + y < 2y < 2z.
Hence 2z > x + y.
***
The problem with this "proof" is that it is unclear and incomplete.

$\text {Proposition } 1:\ x,\ y \in \mathbb Z^+ \text { and } n \in \mathbb N^+ \text { and } x^n + y^n = z^n \implies\\ z < x + y.$

The OP's proof is a tad informal, but perfectly OK.

$\text {Proposition } 2 :\ x,\ y \in \mathbb Z^+ \text { and } n \in \mathbb N^+ \text { and } x^n + y^n = z^n \implies\\ x + y < 2z.$

I do not get the purported proof.

It starts with $x < y$ due to "symmetry." I think what is meant is that

$x \le y$ can be assumed without loss of generality.

I can buy that.

Next comes "Then y < z." Where in the world does that come from? It is simply asserted rather than proved. (I am not saying it cannot be proved. It just isn't.) Consequently, proposition 2 is not proved. But let's ignore that.

Proposition 3 is not proved either, but let's ignore that too.

We then get the following argument, which I admit I do not understand because it seems to rest on a false proposition, namely that

if z < x + y < 2z, then there does not exist any positive odd integer m > 1 such that x + y divides z^m.

But Bruno has pointed out that

30 < 40 < 60 = 2 * 30 and 40 does divide 30^3 because 40 * 675 = 27000.

Now perhaps that is not the proposition on which the argument relies. So we have either a false proposition or a proposition both unstated and unproved. In either case, the purported proof is beyond pathetic.
Thanks from topsquark

Last edited by JeffM1; July 1st, 2017 at 06:14 PM.
JeffM1 is offline  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
refute



Thread Tools
Display Modes






Copyright © 2017 My Math Forum. All rights reserved.