
Re:     conditions for integer solutions
Posted:
Jan 22, 2011 5:48 PM


On Sat, 22 Jan 2011 06:15:13 0800, Deep wrote:
> The following equation is considered for the given conditions. > > u^2k  2^(k1) = mk^2  1 (1) > > Conditions: u, m are integers, prime k > 2, u > k, m > 0, 2u > > 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^(k1) 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

