June 27th, 2017, 08:48 PM  #1 
Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18  Refute this...
...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. *** 
June 27th, 2017, 08:54 PM  #2 
Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 
That's my story and I'm sticking to it.

June 29th, 2017, 09:39 AM  #3 
Senior Member Joined: May 2013 Posts: 115 Thanks: 10  
June 29th, 2017, 04:06 PM  #4 
Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 
Yes

June 29th, 2017, 04:57 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,808 Thanks: 723 Math Focus: Wibbly wobbly timeywimey stuff. 
Where's that fence when you need it? Dan 
June 29th, 2017, 05:16 PM  #6 
Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18  
June 29th, 2017, 10:21 PM  #7 
Senior Member Joined: May 2013 Posts: 115 Thanks: 10 
Say $\displaystyle x+y=2^35$ and z=2*3*5 x+y is not a factor of z or some power of it Still x+y divides $\displaystyle z^n$ (n>2) I know it's a pain in the ass 
June 30th, 2017, 03:57 AM  #8 
Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18  
June 30th, 2017, 07:24 AM  #9 
Senior Member Joined: May 2013 Posts: 115 Thanks: 10  
July 1st, 2017, 08:19 AM  #10 
Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 
bruno59 wrote... Say $\displaystyle x+y=2^35$ and z=2*3*5 x+y is not a factor of z or some power of it Still x+y divides $\displaystyle z^n$ (n>2) I know it's a pain in the ass *** I'm not certain if you meant to write 2*35 instead of 2^35. If you really meant 2^35 = x + y then that would be quite a bit worse (as you can easily see from what follows), so I will go with x + y = 2*35. Then 2*35 = 70 = x + y. If z = 2*3*5 = 30, then 2z = 60. Now we have 2z > x+ y, or 60 > 70  which is clearly not true. As I said, your example contradicts Claim 2. {And now you should be able to see why I supposed you meant x + y = 2*35 rather than 2^35.} 