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Topic: ----- ----- ----- comparison of two equations
Replies: 2   Last Post: Jan 8, 2014 10:40 PM

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James Waldby

Posts: 308
Registered: 1/27/11
Re: ----- ----- ----- comparison of two equations
Posted: Jan 8, 2014 11:43 AM
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On Wed, 08 Jan 2014 08:11:20 -0800, Deep wrote:

> Consider the following two equations below:
>
> z^p - x^p = 2(m)^2 (1)
>
> z^p - x^p = 2(p^p.m)^2 (2)
>
> Given conditions: z, x are odd integers each > 0 and
> z > x, m is an even integer > 0, prime p > 3
>
> Conjecture: If (1) has no integer solutions then (2) also
> cannot have any integer solutions.


It's straightforward to prove the conjecture if x,z,m,p all
are free. Suppose (2) has a solution (x, z, m = m", p).
Then to solve (1) take m = m' = p * m".

Specifying whether either or both of m and p are free variables
vs. fixed constants is a necessity if "has integer solutions" is
to be well-defined.

--
jiw



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