--- --- Irrational solutions
- From: Deep <deepkdeb@xxxxxxxxx>
- Date: Sat, 2 Feb 2008 18:04:43 -0800 (PST)
Consider the following equation under the given conditions.
R^(1/2) = n^(k-2)[S/T] (1)
where S = m^(k-1) - Am^(k-3) + Bm^(k-5) - .. - k
(2)
T = n^(k-1) - An^(k-3) + Bn^9k-5) - ... ..-
k (3)
mn =
1 (4)
Condition: R is positive rational but not a perfect square.
k is a prime > 3, A, B, .. divisible by k
Assertion: m = u^(1/2) where u is rational but not a perfect square
will satisfy (1)
My argument: k-2 is odd and k-1, k-3, etc are even. Therefore S/T is
rational. Since mn=1
n^(k-2) is irrational of the form q^(1/2) where q is rational but not
a perfect square.
This will make left and right sides of (1) consistent.
Helpful comments about the correctness of the assertion will be
appreciated
.
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