Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.


robersi
Posts:
313
From:
Northern New York
Registered:
8/18/09


Re: n does not divide abc
Posted:
Jul 15, 2014 1:23 AM


On Tuesday, July 15, 2014 12:39:32 AM UTC4, quasi wrote: > robersi730 wrote: > > > > > >Assume a^n + b^n = c^n where a,b,c are non zero integers > > >and n is an odd prime. > > > > > >and the gcd(a,b,c)=1 > > > > > >also assume > > > > > >a^na=0 (mod n^2) > > >b^nb=0 (mod n^2) > > >c^nc=0 (mod n^2) > > > > You can't assume the above congruences. > > > > Your claimed proof of those congruences only works if n _doesn't_ > > divide abc. > > > > quasi
It does work. I showed a+b=0 mod p^2 and of course a^p+b^p=0 mod p^2.
This result can be added to your edited version nicely. As for the clarity of this addition I dunno.
a^p = b^p mod (p^p)
if p divides c
a=b mod p by F Little T
a = b + pq
and not
a = b + p^2q'
*note p does not divide q
binomial expansion yield from the former
a^p = b^p mod (p^2 not p^3)
contradiction. therefore
a + b = 0 mod p^2
and
a^p + b^p = a + b mod p^2
when pc
(p doesn't divide ab *said w.o.l.o.g.)



