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Still no claims, on FLT cont'd
Posted:
Sep 29, 1996 7:07 PM


If anyone cared to read my previous post they may think they found an error at the end, but here's why that's not so.
I put out my alternate statement of FLT as
(x+yz)^n = n(zx)(zy)(x+y)Q where Q is the other stuff
for example,
(x+yz)^3 = 3(zx)(zy)(x+y); Q=1
I showed why neither x,y nor z can be divisible by n, which means that
Q must have an n^{n1} factor.
I ended with the expression
[x+y(zf)]^n = n[(zf)x][(zf)y](x+y)Q(f) + x^n + y^n  (zf)^n
and stated that if (zf) is divisible by n then Q(f) must be divisible by n because of Fermat's Little Theorem.
But, of course, there's already an n multiplied times the first term on the right, so I should have further explained that with
the reader can verify that [x+y(zf)]^n  [x^n + y^n  (zf)^n]
must be divisible by n^2. Easiest way is to just use a small test n like n=3.
Now, I can also write it as [(xd)+(ye)z]^n where d+e=f and (xd) and (ye) are divisible by n.
Notice then that my first term on the right is divisible by n^2.
If I subtract (d+e)^n = f^n from both sides I have this on the right
d^n + e^n  f^n with the other terms.
But, unlike the previous case, it is only divisible by n, which is easy to verify.
But the left side is divisible by n^2 and so is the first term on the right. So this is more direct than what I left on my previous post.
I still make no public claims.
James Harris



