Prove Congruence: Prime p|/a & Additive Order of a Modulo p = p

[kathrynag]

Homework Statement

prove that if p is a prime number and a is any integer p|/a(p does not divide a), then the additive order of a modulo p is equal to p.

Homework Equations

The Attempt at a Solution

I know p|/ a says a\neqpn for an integer n.
The additive order of a modulo n is the smallest positive solution to ax\equiv0 mod n.
Let p be a prime number and p|/ a.
Then we can say (p, a)=1. That is p and a are relatively prime.
That's as far as I got.

 

[Office_Shredder]

ax=0 mod p means that p|ax. If p does not divide a, what can you infer?

 

[kathrynag]

the additive order is p?

 

[Office_Shredder]

Why?

 

[kathrynag]

Since p does not divide a, there are no multiples of a that equal p. Thus, p must be the smallest additive order.

 

[Office_Shredder]

Since p does not divide a, there are no multiples of a that equal p. Thus, p must be the smallest additive order.

p not dividing a means no multiple of p equals a, not that no multiple of a equals p. And I don't see what that has to do with additive order anyway... 4 does not divide six, there are no multiples of 4 or 6 that give the other one, but the additive order of 4 mod 6 is three, not six.

If p|ax and p does not divide a, and p is a prime, what must p divide? This is the defining property of prime numbers

 

[kathrynag]

p divides x or p divides 1

 

[Office_Shredder]

p can't divide 1...

You should be able to finish the proof now

 

[kathrynag]

Ok I think this makes a bit more sense for em now. Thanks!