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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+yz)^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+yz)^3 = 3(zx)(zy)(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+yz)^5 = 5(zx)(zy)(x+y)[z^2  (x+y)z + x^2 + xy + y^2]
In general, I just use
(x+yz)^n = n(zx)(zy)(x+y)Q
For other primes other than n, I also have
(x+yz)^a = a(zx)(zy)(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 (zx), (zy) 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(zx)(zy)(x+y)Q(a) term by 3^3 as well. But, if for instance, (zx)
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^{n1}
which didn't seem to lead to anything but I think I found something of interest.
If I write
[x+y(zf)]^n = n[(zf)x][(zf)y](x+y)Q(f) + x^n + y^n  (zf)^n
I find that because of Fermat's Little Theorem, my first terms on the right have to be divisible by n.
If (zf) 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 (zf) 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+yz)^5, you notice that Q can't be divisible by 5.
That's enough typing for one sitting. I'm done.



