July 1st, 2017, 08:30 AM  #11 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
It was 2³5 (which is 40), not 2*35.

July 1st, 2017, 05:45 PM  #12  
Senior Member Joined: May 2016 From: USA Posts: 786 Thanks: 312  Quote:
$\text {Proposition } 1:\ x,\ y \in \mathbb Z^+ \text { and } n \in \mathbb N^+ \text { and } x^n + y^n = z^n \implies\\ z < x + y.$ 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. Last edited by JeffM1; July 1st, 2017 at 06:14 PM.  