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Re: Decomposition of a 10th degree equation
Posted:
Mar 15, 2013 8:30 PM


On Fri, 15 Mar 2013 07:20:46 0700, Deep wrote: > On Friday, March 15, 2013 7:32:06 AM UTC4, Deep wrote: >> Consider the following equation (1) for the given conditions.
>> x^10 + y^10 = z^10 (1) >> >> Conditions: x, z are odd integers > 0 and y is non integer >> but x^10, y^10, z^10 are all integers each > 0 >> >> (1) can be decomposed as (2) and (3) where x = uv and >> u, v are co prime integers. >> z^5 + y^5 = u^10 (2) >> z^5  y^5 = v^10 (3) >> >> It is seen that if (2) and (3) are multiplied (1) is obtained. >> >> Question: Is the decomposition of (1) into (2) and (3) valid? >> >> If not why not. > *** *** *** > KIndly clarify. The solutions of (2) and (3) are also the solutions > of (1). Given, none of (1), (2), (3) has integer solutions. That > is y is a non integer in all of them. KIndly clarify the meanings > of "valid), "invalid" > *** *** ***
Deep, you seem to be using some misconfigured mail software that inserted numerous spurious blank lines, and also you didn't quote the post you replied to, which apparently is following from what Pubkeybreaker wrote on Fri, 15 Mar 2013 at 06:09:29 0700:
> (A) If (2) and (3) are true, then (1) is true. > > This is correct. > > But the converse need not be. The converse says: > > (B) If (1) is true, then (2) and (3) are true. > > And this is an invalid conclusion.
As he points out, if you have (2) and (3) then you have (1).
But in general, when (1) holds and y^10 is an integer, y^5 need not be an integer, so (2) and (3) cannot be satisfied.
For an example, run the following Python code:
x=15; z=17; y = (z**10  x**10)**(1/10.) print x, z; print y, y**5
It prints out: 15 17 16.4367687616 1199726.43124
 jiw



