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Topic: Another try on FLT with no claim that it's worth anything.
Replies: 0

 James Harris Posts: 16 Registered: 12/12/04
Another try on FLT with no claim that it's worth anything.
Posted: Sep 29, 1996 6:05 PM

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.