Re: How to prove this congruence problem???
- In other words, Prove the following false: for any and all p and a,
there always exists an x such that x^3 -a is divisible by p
As Satoshi implied (although mixing up the 'a' and the 'x' I think),
by Fermats Little Theorem, if p = 7, 'a' must be of the form 7u, 7u+1
or 7u-1. Clearly this contradicts the original condition that 'a'
can be any number.
Now on to disproving the Riemann hypothesis, hehe!
--- In firstname.lastname@example.org, Satoshi Tomabechi
> Consider p=7.wrote:
> Fermat's little theorem claims that a^6 = 1 mod 7.
> This implies that a^3 = 1 or -1 mod p for all a.
> Satoshi Tomabechi
> On Thu, 05 Jun 2003 03:20:55 -0000
> "Bill" <iampoliceman@y...> wrote:
> > it should be "for all a". thanks
> > --- In email@example.com, "Bill" <iampoliceman@y...>
> > > Prove it's false.
> > >
> > > for all p (p is prime implies for a there exists an x such that
> > (x^3
> > > is congurence to a (mod p))
> > >
> > > thanks a lot!