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June 27th, 2017, 09:48 PM   #1
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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.
***
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June 27th, 2017, 09:54 PM   #2
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That's my story and I'm sticking to it.
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June 29th, 2017, 10:39 AM   #3
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Quote:
Originally Posted by uvkajed View Post
therefore
x + y can not divide z^n
Νο
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June 29th, 2017, 05:06 PM   #4
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Yes
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June 29th, 2017, 05:57 PM   #5
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Math Focus: Wibbly wobbly timey-wimey stuff.
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June 29th, 2017, 06:16 PM   #6
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Quote:
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Where's that fence when you need it?

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Nice
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June 29th, 2017, 11:21 PM   #7
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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
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June 30th, 2017, 04:57 AM   #8
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Quote:
Originally Posted by bruno59 View Post
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

Your example contradicts Claim 2. You can't do that.
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June 30th, 2017, 08:24 AM   #9
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Quote:
Originally Posted by uvkajed View Post
Your example contradicts Claim 2. You can't do that.
Claim 2 says 2z>x+y and that's exactly what happens in the counter example.
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July 1st, 2017, 09:19 AM   #10
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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.}
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