## How to prove this congruence problem???

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• 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!
Message 1 of 4 , Jun 4, 2003
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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!
• it should be for all a . thanks ... (x^3
Message 2 of 4 , Jun 4, 2003
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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!
• 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
Message 3 of 4 , Jun 4, 2003
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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@...> 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!
>
• 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
Message 4 of 4 , Jun 4, 2003
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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

<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!
> >
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