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Topic:
Can a^p + b^p == c^p mod p^k (odd prime p, 0<a,b,c<p) ?
Replies:
54
Last Post:
May 6, 2002 11:25 AM




Re: Can a^p + b^p == c^p mod p^k (odd prime p, 0<a,b,c<p) ?
Posted:
Apr 30, 2002 1:59 PM


Joe Number <n@ion.isotope> wrote > > Nico Benschop <n.benschop@chello.nl> wrote: > > >> a^p + b^p == c^p mod p^k (odd prime p, 0<a,b,c< p) > >> does have solutions, and normalize to c = p1 > > Sorry I don't have the info. Why can't your PC handle larger p? > What sort of PC do you think could handle the problem? It does > not look like it should require much memory or much speed.
It's just that higher precision, say 64 bits or more, is needed for arithmetic mod p^4 at the parger primes p, say p>10^4. Complete search for a+b=p1, with a<(p1)/2, grows quadratically with p, so full search requires a rather powerful machine, if no counter example is found early on, for smallish p.
> It seems to me if you normalize c to p1 modulo p^k, you will > change the values of a and b so that if a, b, c < p then > a and b might become > p. That is, if normalize means what > I think it means. Well, if they were integers you would be right, but they are residues (mod p^k). For instance [1] has all terms coprime to p, so they are in the (cyclic) group of units, so each residue has a unique inverse so that [1] can be multiplied throughout by the inverse of one of the terms to make it == 1 mod p^k, or here: divide by c^p to get normal form x^p + y^p == 1 mod p^k, or: a 'scale factor' that makes r^p + s^p == (p1)^p mod p^k. considering only pos.integers 0 < r, s, p1 < p
> I read some of the things at your site, and I thought it was > interesting as far as I could understand it. But I am a genuine > amateur and it is a bit complex and the meaning is often unclear. > For instance, what does "core" mean?
I use the simplest principles of residue arithmetic mod p^k (odd prime p), which Fermat around 1640 discovered for case k=1: Fermat Small Thm (FST): n^p == n mod p (odd prime p, all n).
And residue arithmetic was formalized in general by Gauss (1801), as satisfying the same syntax rules as integer arithmetic: associative and commutative (+) and (.) (.) distributes over (+), and (^) over (.), but (^) not over (+)
> For instance, what does "core" mean?
There are p^{k1} residues mod p^k which are a multiple of p, hence 0 mod p, and the same number of residues that are 1 mod p.
All p^k  p^{k1} = (p1)p^{k1} residues mod p^k coprime to p are called 'units' (mod p^k), and they form a cyclic group: one generator suffices to get G_k == \{ g^i \} for i=1 .. (p1)p^{k1}.
This product order of two coprime factors makes G_k == A_k . B_k, where 'core' A_k has order p1 (note: an extention of FST for k>1), and 'extension group' B_k has order p^{k1}, generated by p+1, so B_k == \{ (p+1)^i, for all i=1 .. p^{k1} \}, consisting of all residues mod p^k that are 1 mod p. With G_k being cyclic, also factor groups A_k and B_k are cyclic.
And in cyclic group A_k(.) of order p1 holds: n^p == n for all n in A_k (i.e. for each core residue n).
Re: FLT for integers .vs. for residues mod p^k: Each solution of: a^p + b^p == c^p "in core" A_k has:
a^p==a, b^p==b and c^p == (a+b)^p == a+b == a^p + b^p (mod p^k)
Hence: exponent p distributes over a sum (EDS property) which "blocks extension to integers", meaning: integers A,B,C with A^p==a^p, B^p==b^p, C^p==c^p (mod p^k) cannot exist [since EDS does not hold for integers].
> Caeo, Nico > > PS: Hey, are you Italian, Dutch, or English?
I'm Dutch (see CV on my homepage;) I just, sometimes, use Ciao for fun.
 NB  http://home.iae.nl/users/benschop/scimat98.htm http://home.iae.nl/users/benschop/sgrpflt.htm http://home.iae.nl/users/benschop/carry.htm



