Fermat's Last theorem

When I first read about this theorem, I thought I had an solution to prove it wrong. For all numbers (n, -n, 0) satisfy the eqtuation. But after reading on Wikipedia, I got a clear picture of what the theorem exactly is. :). Here is the definition from Wikipedia link:

In number theory, Fermat's theorem states that no three positive integers a,b, and c can satisfy the equation a^n + b^n = c^n for any integer value of n greater than 2.

So, 0 is out of question. From Wikipedia, it seems Fermat proved the theorem indirectly through technique of infinite descent for n=4. I have no clue what this technique is. (maybe I should read this one soon) I tried solving Fermat theorem, using my crude basic mathematical methods for n=3. Here is my proof (?) ;)

Let the 3 numbers be a,b,c such that a less than b less than c
It is trivial to prove for cases a + b less than c and a + b = c

Let us assume a, b and c do not have any common prime factors. (Proof in next post)

For third case, a + b > c, let us assume a + b = cx.
c > x > 1
This leads that a^2 - ab + b^2 = cy.
c = xy
a + b and a^2 - ab + b^2 must be integers, which are product of factors of c.
So, x may not be an integer, but must be an rational number, which can be given in form m / n, where m > n and m and n are product of some of the factors of c.
a + b = cx, x greater than 1

a^2 + 2ab + b^2 = c^2*x^2
a^2 -ab + b^2 = cy
3ab = c^2 (x^2 - 1/x)
Since, 3, a, b and c are integers, x^2 - 1/x should be an integer (Well, original proof was based on equality, but it can be only applied if x > 2, and too tired to think about it now, maybe some time later...). Since x^2 - 1/x can never be an integer for any rational number x, greater than 1, it follows that a^3 + b^3 can never be equal to c^3.


Phew! Comments welcome :)

1 Rupee = 1 Paise???

Here is the calculation...
Trying to find the error in it.
1 Rupee = 100 paise
= 10 paise * 10 paise
= 0.1 Rs * 0.1 Rs
= 0.01 Rs
= 1 paisa

Still wondering... Only flaw I can think of is unit conversion not allowed.