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Topic: -- FLTcase1 by finite descent with Hensel's lift --
Replies: 4   Last Post: Aug 25, 1998 4:52 PM

 Messages: [ Previous | Next ]
 Nico Benschop Posts: 551 Registered: 12/4/04
Re: -- FLTcase1 by finite descent with Hensel's lift --
Posted: Aug 19, 1998 3:25 AM

Kent & Kat wrote:
>
> Yes, my post was in error - I glossed too quickly over the definition of
> "core A_k. " I don't get the 43, though. The 7th power residues mod 49 are
> {1, 30, 31, 18, 19,48}, and 3^7 = 31 mod 49, 5^7 = 19 mod 49.

Note my using base7 code: 31(dec) = 4.7+3 = 43(base7)
This base_p code is handy, because A_{k+1} = A_k mod p^k is then
obvious:
the higher order core only differs in the msd (most signif digit).

> In any case, here we do have x^7 = x = 31 mod 49, y^7 = y = 19 mod 49, and
> z^7 = z = 1 mod 49. So the equation x^7 + y^7 = z^7 mod 49 reduces to x + y
> = z mod 49, as you assert.
> However, your claim is that we consider the equation with numbers
> instead of residues, x + y = z implies x^7 + y^7 < z^7. The problem is we
> don't have x + y = z, as the above example shows, we only have x + y = z mod
> 49. In fact, in the above case, with numbers we get the opposite inequality
> : x^7 + y^7 > z^7.
> Doesn't this invalidate your argument?
> Kent.

No, the whole exercise shows that if you ignore carries,
that is: beyond weight p^{k-1}, you *can* get equality, but *only*
at the cost of *inequality* for integers (that is: _including_ carries).
That is the whole point of the proof. I would really advise to read the
paper first: ref[1] on my homepage. Or better still: take ref[5] which
treats aspects (cubic roots of unity mod p^k) that Fermat easily might
have found himself (causing his enthusiastic marginal remark ?)
"On Fermat's marginal note: a suggestion".

Enjoy.
--
Ciao, Nico Benschop. | If stuck@closure (mod...) use the carry
http://www.iae.nl/users/benschop | xxxxxxxxxxxxxxxx.xxxxxxxxxxxxxxxx

Date Subject Author
8/19/98 Nico Benschop
8/19/98 Kent
8/20/98 Nico Benschop
8/25/98 Nico Benschop