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 July 1st, 2017, 08:30 AM #11 Global Moderator   Joined: Dec 2006 Posts: 19,168 Thanks: 1640 It was 2³5 (which is 40), not 2*35.
July 1st, 2017, 05:45 PM   #12
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Quote:
 Originally Posted by uvkajed ...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.

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.

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