Re: ----- ----- Prime number
- From: david <dlee753@xxxxxxxxx>
- Date: Sun, 3 Aug 2008 17:15:09 -0700 (PDT)
On Aug 3, 9:25 am, Deep <deepk...@xxxxxxxxx> wrote:
Consider the following equation (1) under the given conditions.****
k^(mk-1) = [(u^k).2ab]/[(v^k).(a^2 - b^2)] (1)
Conditions:
(1) u, v, a, b are all integers each > 1, integer m > 0.
(2) v is even and u is odd, b is even and a is odd, k is a prime > 5.
(3) (a, b) = 1, (u, v) = 1; (2ab, (a^2 - b^2)) = 1; ( ) = 1 means
quantities are relatively prime
Assertion: (1) does not have any solution.
Any comment about the correctness of the assertion will be highly
appreciated.
only possible solutions are:
a = k^(mk-1), a^2 - b^2 = u^k, 2b = v^k. Otherwise, the given equation
is inconsistent.
Any comment will be appreciated.
****
.
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