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




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^{k1}, 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



