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Topic: ---- ---- ----- ----- conditions for integer solutions
Replies: 3   Last Post: Jan 23, 2011 7:21 PM

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Calumnist

Posts: 554
Registered: 12/11/04
Re: ---- ---- ----- ----- conditions for integer solutions
Posted: Jan 22, 2011 5:48 PM
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On Sat, 22 Jan 2011 06:15:13 -0800, Deep wrote:

> The following equation is considered for the given conditions.
>
> u^2k - 2^(k-1) = mk^2 - 1 (1)
>
> Conditions: u, m are integers, prime k > 2, u > k, m > 0, 2|u
>
> Question: Under the given conditions can (1) be satisfied?
>
> My observation: Since the LHS of (1) is even the RHS of (1) must
> also be even. This implies m must be odd. Since k is odd the
> LHS is the difference of two squares.


If the difference you refer to is the difference of u^2k and
2^(k-1) then LHS is not the difference of two squares, because
u^2k cannot be a square, as follows: Suppose odd prime k divides
u; then k appears to an odd power (ie, 3, 5, ...) in the product
u^2k. Suppose k does not divide u; then k appears to an odd power
(ie, 1) in the product u^2k. Hence u^2k is not a square.

[ignored rest]
> Therefore, the RHS of (1) cannot be 2mod(4). Otherwise the LHS can not
> be a difference of two squares.

...

--
jiw




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