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!

Mark

--- In

primenumbers@yahoogroups.com, Satoshi Tomabechi

<tomabeti@c...> wrote:

> Consider p=7.

> 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 primenumbers@yahoogroups.com, "Bill" <iampoliceman@y...>

wrote:

> > > 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!

> >