Thread: Multiplicative order

Hi, need a some help with this proof.

Prove the multiplicative order of is for odd prime .

How do I go about this?

Thank You for any help.

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

step one: what is ? justify your answer.

Originally Posted by Deveno

step one: what is ? justify your answer.

I am sorry, I don't know.

I can see that does not divide .





which is false for positive n.

I need a little more help if you can.

But thanks for your reply.

try expanding using the binomial theorem. which terms don't have in them?

Originally Posted by Deveno

try expanding using the binomial theorem. which terms don't have in them?

So in the expansion of the only terms than don't have in them are and .

So where to from here then?

Thanks for your help Dev

um, no....

the coefficent of p in the expansion is p:

where all the remaining terms involve higher powers of p.

thus .

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?

Originally Posted by Deveno

um, no....

the coefficent of p in the expansion is p:

where all the remaining terms involve higher powers of p.

thus .

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.

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".

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.