Re: --- --- --- Solutions of an equation
- From: quasi <quasi@xxxxxxxx>
- Date: Sat, 25 Oct 2008 18:03:34 -0400
On Sat, 25 Oct 2008 09:17:49 -0700 (PDT), Deep <deepkdeb@xxxxxxxxx>
wrote:
Assertion:
The following equation (1) cannot be satisfied if all of x, y, z are
integers each > 1.
x^2 + 8y^2 = z^2 (1)
Your assertion is false.
Try, x=2, y=2, x=6.
Another solution: x=41, y=15, z=50.
My approach: x, (sqrt (8)y), z represent the three sides of a
Pythegorian triangle. Given y is an integer sqrt(8)y can never be an
integer.
Right -- since y is a positive integer, sqrt(8)y is irrational, hence
is not an integer. So what?
Therefore, it can be argued that (1) has no integer solutions.
No, the logic is flawed -- you haven't shown a contradiction.
Moreover, since solutions actually exist, a contradiction is not
achievable.
quasi
.
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