1. ## Multiplicative order

Hi, need a some help with this proof.

Prove the multiplicative order of $(1+n) \ mod \ n^2$ is $n$ for odd prime $n$.

Thank You for any help.

2. ## Re: Multiplicative order

You have to try first. What is the definition of the multiplicative order?

3. ## Re: Multiplicative order

step one: what is $(1 + p)^p \ (mod \ p^2)$? justify your answer.

4. ## Re: Multiplicative order

Originally Posted by Deveno
step one: what is $(1 + p)^p \ (mod \ p^2)$? justify your answer.
I am sorry, I don't know.

I can see that $n^2$ does not divide $(1+n)^n$.

$\frac{(1+n)^n}{n^2}=0$

$(1+n)^n=0$

which is false for positive n.

I need a little more help if you can.

5. ## Re: Multiplicative order

try expanding $(1+p)^p$ using the binomial theorem. which terms don't have $p^2$ in them?

6. ## Re: Multiplicative order

Originally Posted by Deveno
try expanding $(1+p)^p$ using the binomial theorem. which terms don't have $p^2$ in them?
$(p+1)^p= \sum_{k=0}^p \tbinom pk p^k \cdot 1^{p-k} = \sum_{k=0}^p \tbinom pk p^k$

So in the expansion of $(1+p)^p$ the only terms than don't have $p^2$ in them are $1$ and $p$.

So where to from here then?

7. ## Re: Multiplicative order

um, no....

the coefficent of p in the expansion is p:

$(p+1)^p = 1 + p^2 + \frac{p(p-1)}{2}p^2 + \dots$ where all the remaining terms involve higher powers of p.

thus $(p+1)^p = 1 (mod\ p^2)$.

so the multiplicative order of p+1 (mod p^2) has to divide p. what are our choices, given that p is an odd prime?

8. ## Re: Multiplicative order

Originally Posted by Deveno
um, no....

the coefficent of p in the expansion is p:

$(p+1)^p = 1 + p^2 + \frac{p(p-1)}{2}p^2 + \dots$ where all the remaining terms involve higher powers of p.

thus $(p+1)^p = 1 (mod\ p^2)$.

so the multiplicative order of p+1 (mod p^2) has to divide p. what are our choices, given that p is an odd prime?
Oh yeah, slight error on my part. So if the order of p+1 mod p^2 has to divide p and p is an odd prime, then by the definition of a prime, the order of p+1 mod p^2 has to be p.

9. ## Re: Multiplicative order

almost...why isn't the order of p+1, 1 (after all, 1 divides p)?

it's a small detail, and easily answered, but it never hurts to dot the "i's" and cross the "t's".

10. ## Re: Multiplicative order

Originally Posted by Deveno
almost...why isn't the order of p+1, 1 (after all, 1 divides p)?

it's a small detail, and easily answered, but it never hurts to dot the "i's" and cross the "t's".
Yes, of course! That makes perfect sense. Thanks again for your fantastic help.