# I Some number theory proofs

1. Mar 16, 2016

### cragar

I just want to make sure I understand these number theory proofs.
$b^{\phi (n)}=1mod(n)$
$\phi (n)$ is the order of the group, so b to some power will equal the identity. so thats why it is equal to one. $b^p=bmod(p)$
$b^p=b^{p-1}b$
$b^{p-1}$ produces the identity since p-1 is the order of the group. so thats why it equals b.

2. Mar 16, 2016

### FactChecker

Can you clarify what statement you are trying to prove and which statements are the proof? Is b any arbitrary element of the group? Are you trying to show that p = Φ(n)? What is n?

3. Mar 16, 2016

### micromass

Staff Emeritus
And what group are you talking about?

I think you're trying to prove Euler and Fermat's little theorem somehow, but then you'll need to put in some more details...

4. Mar 16, 2016

### cragar

Its the multiplicative group mod n , n is a natural number. now b is an element of the group. so b to power of the order of the group will contain the order of b.
that is $b^m=e$ since m is contained in the order of the group,we get the identity. This proves $b^{\phi (n)}=1mod (n)$.
the second one $b^p=b mod (p) = b^{p-1}b=b^{\phi (p)}b= b mod (p)$
I will prove that the order of an element exists. sine b is an element of the group then we can take powers of b like $b^2 , b^3, .... b^n ...$ now for some m and n and m not equal to n,since our group is finite.
we get $b^m=b^n$ now we multiply both sides by $b^{-n}$
so now $b^{m-n}=e$.
m-n is the order of b.

Last edited: Mar 16, 2016