I'll illustrate with a few examples.

And so on.

]]>Will you please explain me post #12 once again?

]]>Proof: Take *b* = *a*[sup]*p*−2[/sup].

Moreover, by the division algorithm, we are always guaranteed to find a unique *b* in the range 1 ≤ *b* ≤ *p*−1 such that *a*[sup]*p*−2[/sup] ≡ b (mod *p*).

In the language of group theory, we say that *b* is a multiplicative inverse of *a* modulo *p*. This is the result I was using in my proof.

Are you familiar with Fermat's little theorem? This states that if

May be I didn't understand it]]>

The set , where is prime, forms a group under multiplication modulo .]]>

Suppose is prime. We first note that .

Consider the multiplicative group of

modulo , where is odd. We have and . Conversely, if and , then divides divides or or . Hence only 1 and are their own multiplicative inverses modulo .It follows that

. (Each number in the list multiplies by another number in the list to ≡ 1 (modConversely suppose

and . Then for some integer . Let be a prime divisor of . If were not prime, would be smaller than and so would divide and hence would divide , a contradiction. Thus must be prime.]]>The way to prove that is divisible by for composite is as follows.

If

is composite and greater than 4, then can be written as a product of an integer greater or equal to 3 and an integer greater or equal to 2, i.e. where and .Now I claim that

. Proof. + rearranging gives as claimed.This means that the factors of

contain a sequence of consecutive integers and a nonoverlapping sequence of integers. The first sequence contains a factor of and the second sequence contains of factor of . So the product of those integers is divisible by . But the two sequences of integers are nonoverlapping and so is divisible by the product of those integers. Hence is divisible by .]]>a good bit of number theory. I am not a number theorist so it refers to things I am not familiar with.

That's why I look at the problem the way I did in my previous post. Maybe someone else can explain

the number theory congruence bit for you.

Good luck!

]]>