Re: --- --- --- Solutions of an equation
- From: Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Tue, 28 Oct 2008 02:27:21 GMT
In article
<a3444ad0-0bfd-4cbe-8dbc-45989b9d1d32@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Deep <deepkdeb@xxxxxxxxx> wrote:
x^2 + 8y^2 = z^2 (1)
Try, x=2, y=2, x=6.
[presumably, that should be z = 6]
Another solution: x=41, y=15, z=50.
Typo: In the above line, it should be z = 59.
**** A general solution of x^2 + 8y^2 = z^2 (1) can be obtained as
(2*sqrt(2)y)^2 + (y^2 - 2)^2 = (y^2 + 2)^2 (2)
Therefore, z = y^2 + 2, x = y^2 - 2. From this one gets z - x =
4 (3)
Apparently, (3) does not satisfy all the above examples.
You may kindly offer sone explanation. Thanks
*****
8 y^2 = z^2 - x^2 = (z - x) (z + x)
z - x = 2 m^2, z + x = 4 n^2
x = 2 n^2 - m^2, y = m n, z = 2 n^2 + m^2
gives a two-parameter family of solutions. There may
be other, similar, families, based on small variations
of the calculations above.
m = 3, n = 5 gives x = 41, y = 15, z = 59.
--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.
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