> I should specify that I am looking for the first value of x other than

In that case, the smallest counterexample is for modulus 45.

> 1. So in your example the first x that fits the requirements is 4.

The smallest x>1 such that x^2 == 1 (mod 45) is x=19.- Hi Jack,

Thanks for that. I must admit that I wasn't looking at low numbers,

but larger bi-primes.

I'll have to revisit my stratgey to handle composites which are the

product of more than two (not necessarily unique) primes.

Regards,

Kevin.

>

than

> > I should specify that I am looking for the first value of x other

> > 1. So in your example the first x that fits the requirements is

4.

>

->

> In that case, the smallest counterexample is for modulus 45.

>

> The smallest x>1 such that x^2 == 1 (mod 45) is x=19.

>

>

>

>

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> - Hi Jens,

I think that my point is that every odd composite has an even 'x' where

x^2 = 1 mod p. In the case of p = 55 then x is 34.

So what I am getting at, is that every odd composite will have a case where

x is even (and less than p-1), but may not have a case where x is odd.

The counter example that I need to disprove this is where an odd composite

(p) does not conform to the statement x^2 = 1 mod p where x is less than p-1.

Regards,

Kevin.

At 07:53 PM 7/10/2004, you wrote:>----- Original Message -----

>From: "Kevin Acres" <research@...>

>To: "Jack Brennen" <jack@...>; "Primenumbers"

><primenumbers@yahoogroups.com>

>Sent: Thursday, October 07, 2004 3:53 AM

>Subject: Re: [PrimeNumbers] Modulo 1 question

>

>

> > Hi Jack,

> >

> > Thanks for that. I must admit that I wasn't looking at low numbers,

> > but larger bi-primes.

> >

> > I'll have to revisit my stratgey to handle composites which are the

> > product of more than two (not necessarily unique) primes.

> >

> > Regards,

> >

> > Kevin.

>

>Smallest bi-prime counter example:

>21^2 == 1 (mod 55)

>

>If you can program in any language, try experimenting before posting.

>

>--

>Jens Kruse Andersen (offline) - --- Kevin Acres wrote:
> So what I am getting at, is that every odd composite will have

Okay, let's look at this a couple of things at a time...

> a case where x is even (and less than p-1), but may not have a

> case where x is odd.

First, I'll change the modulus from 'p' to 'm' because 'p' is

usually reserved for primes:

x^2 == 1 (modulo m) {m is an odd composite}

Note that if there exists any solution x, with 1 < x < m-1, then

there exist two such solutions. This is easy to see, because

if we have:

a^2 == 1 (modulo m)

then x=(m-a) is also a solution:

(m-a)^2 == m^2-2*a*m+a^2 == m*(m-2*a)+a^2

m*(m-2*a)+a^2 == a^2 (modulo m)

m*(m-2*a)+a^2 == 1 (modulo m)

Now, since m is odd, the pair (a,m-a) contains both an odd

number and an even number.

So, we can restate: If there exists any solution x, with

1 < x < m-1, then there exist both odd and even solutions

which satisfy that constraint.

Assume that m is a product of two odd numbers j and k which

are greater than one and which are coprime.

To solve x^2 == 1 (modulo m), we can solve:

u^2 == 1 (modulo j)

v^2 == 1 (modulo k)

and then by the Chinese Remainder Theorem, each unique pair

of solutions (u,v) leads to a unique solution x. There are

at least 4 solutions to this system of modular equivalences:

(u,v) == (1,1)

(u,v) == (1,k-1)

(u,v) == (j-1,1)

(u,v) == (j-1,k-1)

The first of these four leads to the solution x == 1, and the

last leads to the solution x == m-1. The other two also lead

to solutions to the original equation, and the Chinese Remainder

Theorem states that those solutions will fall in the range:

0 <= x <= m-1, and that they will be distinct from the other

solutions. A little thought shows that these two solutions are

of the form x==a and x==(m-a) as discussed earlier, and that

the existence of both odd and even solutions is directly shown.

The only remaining problem is in the case where m cannot be

expressed as a product of coprime odd numbers > 1. If m is

an odd composite, the only way this can happen is if m is a

perfect power of a prime.

I won't go into the proof by induction that x^2 == 1 (mod p^n)

has only the two trivial solutions x == 1 and x == p^n-1, but

it's not too hard to find.

In any case, the final conclusion is that:

x^2 == 1 (mod m) {m an odd number}

has both odd and even non-trivial solutions (1 < x < m-1)

if and only if the modulus m is divisible by at least two

distinct primes.

Hope that helps.