From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: solving FLT_1
Date: 29 Oct 1999 19:30:04 GMT
Newsgroups: sci.math
Keywords: Counterexample to Fermat's Last Theorem for exponent 7 (7-adic)

In article <3819956C.235FDCB1@iae.nl>, Nico Benschop  <benschop@iae.nl> wrote:
>Not so much for your benefit, but rather for possibly other readers,
>I'd like to explain this "carry"-thing with an example 

>2.  Now comes the (non-residue) second step, to go to integers:
I haven't the patience to attempt to decipher what, exactly, is being
claimed in this step, but the example is perhaps unrepresentative:

>The smallest specific example is: a^3 = 1 mod 7^2, the cubic roots of
>unity, with x = a       = 42 (base 7, that is: 4.7+2=30 dec)
>        and y = a^{-1}  = 24
>         + ------------------    
>      and x+y = a + 1/a = 66 = -1 (mod 7^2)
>
>where for integer p-th powers:   X^7 =  1402646634642 
>                                 Y^7 =    21112533024
>                             + ------------------------
>                                        1434063500666 
>                                                |||||
>                       while (X+Y)^7 = 60262046400666  (2.7=14 digits)
>
>which are only equal mod 7^5, 
>---> and the remaining 14-5=11 carries make the difference, see?

I happen to have a counterexample to Fermat's Last Theorem, Case 1, 
exponent 7. Rather than reveal it to the world, let me just say the
numbers are very large and, in keeping with the poster's preference,
I will show the end of the base-7 representations of these numbers:

X   =  ...0000000000000001
Y   =  ...0000000000000024
Z   =  ...6353054001000025

X^7 =  ...0000000000000001
Y^7 =  ...0000021112533024
Z^7 =  ...0000021112533025

Actually the strings of zeros in the expansions of  X  and  Y  are
much much longer than shown, so that  X^7  and  Y^7  have extremely
long strings of zeros to the left of the portion shown, which is why
I didn't show any more digits. The only thing I guess I need to 
convince you of is that there is in fact a number  Z  whose 7th power
also has that many zeros in its expansion, followed by the 21112533025.
This will show that there is no reason to expect "carries" which "make
a difference" any time soon...

Of course, if you think you can prove FLT, you ought to be able to
explain why you think I am mistaken (or lying); I just don't see how
your arguments about "carries" will carry any weight.

Not, of course, that I'm particularly interested in at attempt to
string together some verbiage of the sort we've already seen.

dave
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[Minor typo in original post corrected above --djr]
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