Harristotelian Logic

I probably embarrassed myself enough last time but I've come across some more curious stuff that seems to relate to FLT. I figured it wouldn't hurt to put it out here.

All variables are Naturals.

Consider a^2 + b^2 = c^2 and of course x^n + y^n = z^n

where a^2 > z^n and b^2 > z^n

Since all terms of the form j^n are summations of odds the above means that the squared equation contains the terms of the other one.

For example: j^3 = the summation from 1 to j of 3k^2 - 3k + 1

I need to introduce a new term which I call the square equivalent (SE)

SE equals the square which would have any given odd as its last term.

For example: With 2^3 the last odd is 3(2^2) - 3(2) + 1 = 7

Then SE(2^3)=(7+1)/2=4 which just means that 7 is the last odd term of

the summation that gives 4^2.

Now the rest,

a^2 - x^n + b^2 - y^n = c^2 - z^n

Because summation terms are squared, I can rewrite the above in the following way.

a^2 - x^n + b^2 - SE(y^n)^2 = c^2 - SE(z^n)^2 + Q

Q equals the summation from b+1 to SE(z^n) of 2j-1 - (z^n - y^n)

Putting it all together leads to the following:

SE(z^n)^2 - SE(y^n)^2 = the summation from b+1 to SE(z^n) of 2j-1

Which undoubtably looks strange but here's an example with n=3

SE(z^3)= (3z^2 - 3z)/2 + 1

So the above is

[(3z^2-3z)/2+1]^2 - [(3y^2-3y)/2+1]^2 = the summation from b+1 to (3z^2-3z)/2 + 1) of 2j-1 = [(3z^2 - 3z)/2 + 1]^2 - b^2

Which gives

[(3y^2 - 3y)/2 + 1]^2 = b^2

But I don't stipulate a single value for b^2 and only require that it be greater than z^n along with a^2. And that a^2 + b^2 = c^2.

I hope that's interesting and at least somewhat understandable. It's actually simple if you play with summations a lot.

I make no claims but will appreciate comments. At least I still seem to be original.