Fermat's Last theorem and its consequences

Years ago, French Mathematician Pierre Fermat stated his famous Last Theorem, that says, whereas the equation x^2 + y^2 = z^2 has an infinite no. of solutions, any equation of the type x^n + y^n = z^n where n > 2 has no solution whatsoever.

Many approaches were made, after which Euler was able to construct decent proofs for both n = 3 and n = 4. He was followed by Dirichlet, who constructed another decent proof for n = 5.

At this point, a general approach was made using a well-known property of numbers. It states that the equation a^2 - b^2 = c^2 has no integral solution of its own, which is easy to show from factorization. Thus for any even no. n, the theorem was firmly disproved because, for n=2m the equation can be re-written as (z^m)^2 - (x^m)^2 = (y^m)^2. Thus it only remained to disprove the theorem for odd primes. Further approaches were carried on, it continues even now-a-days.

I myself tried to disprove the theorem for n = 3, by using simple number theory approaches and found that it becomes extremely laborious and difficult. For readers' digest, I present a glimpse of my approach.

We are considered with the equation x^3 + y^3 = z^3. Via factorization, we rewrite the equation as,

x^3 + y^3 = z^3
=> (x + y)(x^2 + y^2 - xy) = z^3

so, the possible cases are:

1) (x + y) = kz
   (x^2 + y^2 - xy) = z^2/k

2) (x + y) = z^2/k
   (x^2 + y^2 - xy) = kz

3) (x + y) = kz^2
   (x^2 + y^2 - xy) = z/k

4) (x + y)=  z/k
   (x^2 + y^2 - xy) = kz^2

since (x^2 + y^2 - xy) > (x + y) for x,y > 2, so the 3rd case is impossible(since kz^2 < z/k so it claims k^2z < 1, a contradiction). We try to carry on with our remaining cases.

So,the first case follows like this,

x + y = kz
=> x^2 + y^2 + 2xy = k^2.z^2
=> 3xy = z^2(k^2 - 1/k) [subtracting the two equations from one another]
=> xy = z^2(k^3 - 1)/3k

To claim that the R.H.S. must be integer, we must have 3k | (k^3 - 1) or 3k | z^2.
But k does not divide (k^3 - 1) so the first possibility is discarded.

Thus, z^2 = 3^(2n).p^2.k^2 [where p contains no more factor of 3]
so, xy = 3^(2n - 1).p^2k.(k^3 - 1)

Finally, (x - y)^2 = (3^n.p.k^2)^2 - 4.3.3^(2n-2).p^2k[k^3-1]
=> (x - y)^2 = 3^(2n - 2).p^2[9k^4 - 12k(k^3 - 1)]

Thus, it reduces to the fact that, we must have k^4 - 12k(k^3 - 1) = some perfect square
or, k(12 - 3k^3) = some perfect square

which is possible only for k = 1, since for k > 1, 12 < 3.k^3. At k = 1 it is (12 - 3) = 9 = a perfect square.
Thus, x + y = z
x^2 + y^2 - xy = z^2 = (x + y)^2 = x^2 + y^2 + 2xy

which leads to xy = 0, a contradiction. Thus such a case is not obtainable in practical.
The other two cases will similarly lead to some sort of contradiction. But I think I already bored you enough of these approaches! So I stop it here...you can surf the net to get lots of modern techniques used to show the results for n=3,5...etc.

I haven't found the proofs constructed by Euler and Dirichlet. If someone knows any link on them, please comment below...