Harristotelian Logic

I've been distracting myself from personal problems by playing around with FLT. Being hardheaded I've both refused to accept that I'm trying the impossible, and have continuously made posts that I had found something only to find out later that I'd made some simple error. But I'm still hardheaded so I keep posting…but no more claims to its worth.

I finally convinced myself that my previous methods weren't going to lead to what I wanted, but because of them I started playing with something else:

(x+y-z)^n with x,y,z relatively prime, n odd prime;

because it held all of the information and was interesting besides.

For instance, if x^3 + y^3 = z^3 then (x+y-z)^3 = 3(z-x)(z-y)(x+y)

which is undoubtably trivial to you but interesting to me.

If I go higher, I find that

x^5 + y^5 = z^5 means that

(x+y-z)^5 = 5(z-x)(z-y)(x+y)[z^2 - (x+y)z + x^2 + xy + y^2]

In general, I just use

(x+y-z)^n = n(z-x)(z-y)(x+y)Q

For other primes other than n, I also have

(x+y-z)^a = a(z-x)(z-y)(x+y)Q(a) + x^a + y^a - z^a

Using the above, there's an easy proof for n=3.

Because x,y or z must be divisible by 3, either (z-x), (z-y) or (x+y) must have a factor of 3^2 with other factors to the third.

If I use my last equation above and since all other odd primes are greater than 3, the left side of the equation is divisible by 3^a and

x^a + y^a - z^a is divisible by 3^3, so I have to be able to divide the

a(z-x)(z-y)(x+y)Q(a) term by 3^3 as well. But, if for instance, (z-x)

has the 3^2 factor, none of the other terms can have a factor of 3, since x,y,and z are relatively prime.

That is also a proof for why neither x,y nor z can be divisible by n, or the same thing would work for them.

However, then it can then be seen that Q is forced to have a factor of

n^{n-1}

which didn't seem to lead to anything but I think I found something of interest.

If I write

[x+y-(z-f)]^n = n[(z-f)-x][(z-f)-y](x+y)Q(f) + x^n + y^n - (z-f)^n

I find that because of Fermat's Little Theorem, my first terms on the right have to be divisible by n.

If (z-f) is divisible by n then that requires that Q(f) be divisible by n. But that puts an interesting requirement on some of the terms within Q(f) and Q.

Especially when it's noted that I can use an f that doesn't make (z-f) divisible by n and doesn't make any of the other terms besides Q(f) divisible by n either.

I guess that's enough. One side note, if you look at the expression for (x+y-z)^5, you notice that Q can't be divisible by 5.

That's enough typing for one sitting. I'm done.

This is a strange one but it works, so what can I say?

Prove x^n + y^n = z^n false when x,y,z relatively prime integers and n a Natural prime greater than 2.

use x - delta x = my, y - delta y = mx, z + delta y = mz

this requires that m = (z+y)/(z+x), delta x = -(y-x)(x+y+z)/(z+x)

delta y = z(y-x)/(z+x)

substituting into the first equation gives

(x - delta x)^n + (y - delta y)^n = (z + delta y)^n

Multiplying out and subtracting off x^n + y^n = z^n , you have all terms multiplied by n except

(delta x)^n + 2(delta y)^n

(I added (delta x)^n to both sides)

But that proves that (x+y+z)^n + 2z^n must be divisible by n

so by F. Little T. x+y+z + 2z must be divisible by n, and therefore

z [x+y-z+x+y+z+2z=2(x+y+z)]must be divisible by n. (yes it's ugly but it works)

Note that z+x and (y-x) can't be divisible by n, which is obvious.

So, just in case you think that I've just managed to prove that z is divisible by n take the following

x - delta x = -mz, y - delta y = my, z = delta y = -mx

do the same thing and you prove that y must be divisible by n.

Well, that does it and I just have to say that the above is ugly. I can understand why a certain someone didn't publish it before. Personally, I'm feeling a bit nauseous.

Signing out.

James S.

do the rest of the above and you prove that y must be divisible by n

Work out everything the same as above and you get that y must be divisible by n.