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Topic: Decomposition of a 10th degree equation
Replies: 4   Last Post: Mar 16, 2013 10:17 AM

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Deep Deb

Posts: 396
Registered: 12/6/04
Re: Decomposition of a 10th degree equation
Posted: Mar 16, 2013 10:17 AM
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On Friday, March 15, 2013 7:32:06 AM UTC-4, Deep wrote:
> Consider the following equation (1) for the given conditions.
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> x^10 + y^10 = z^10 (1)
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> Conditions: x, z are odd integers > 0 and y is non integer but x^10, y^10, z^10 are all integers each > 0
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> (1) can be decomposed as (2) and (3) where x = uv and u, v are co prime integers.
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> z^5 + y^5 = u^10 (2) z^5 - y^5 = v^10 (3)
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> It is seen that if (2) and (3) are multiplied (1) is obtained.
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> Question: Is the decomposition of (1) into (2) and (3) valid?
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> If not why not.
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> Any helpful comment will be appreciated.


***** ***** *****
This posting is reference to the comments by jiw, 3/15/13
(note the computer is not behaving properly)

A simple numerical example may help clarify the matter

Let x^4 -7x^3 + 15x^2 + 5x + 3 = 0 (1)
(1) can be factored as (x^2 -4x + 1)(x^2 -3x +2) = 0.
Therefore, (2) and (3) are obtained.

x^2 -4x + 1 = 0 (2)
x^2 -3x + 2) =0 (3)
From (2): x = 2 +-sqrt(3), from (3): x = 1, 2

NOTE: Anyone of the four(4) roots of (2) and (3) will satisfy (1)
BUT Only two(2) rots of (1) out of four(4) will satisfy either (2) or (3)

THEREFORE, the solutions of (2) and (3) will imply the solutions of (1) BUT the reverse in NOT Correct.
Hope this will clarify the situation.

Thanks for your kind attention, jiw.

Deep




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