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Topic: Still no claims, on FLT cont'd
Replies: 2   Last Post: Sep 29, 1996 11:33 PM

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 James Harris Posts: 16 Registered: 12/12/04
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+y-z)^n = n(z-x)(z-y)(x+y)Q where Q is the other stuff

for example,

(x+y-z)^3 = 3(z-x)(z-y)(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^{n-1} factor.

I ended with the expression

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

and stated that if (z-f) 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-(z-f)]^n - [x^n + y^n - (z-f)^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 [(x-d)+(y-e)-z]^n where d+e=f and (x-d) and
(y-e) 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

Date Subject Author
9/29/96 James Harris
9/29/96 Eduardo Chappa L.
9/29/96 Colin Andrew Percival