Re: --- --- --- Solutions of an equation

On Sat, 25 Oct 2008 18:03:34 -0400, quasi <quasi@xxxxxxxx> wrote:

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.

Typo: In the above line, it should be z = 59.

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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