# Number TheoryEulers phi function, orders, gcd

#### tda120

##### New member
Let a, k , l , m e Z>1 and let a^k=1 (mod m) and a^l= 1 (mod m).
Let d=gcd(k,l)
Prove that a^d=1 (mod m).

I get already confused at the start: Is it true that k|phi(m) (Lagrange) but k can also be a multiple of the order of a (mod m) and then it can be the other way round.

Can anybody clarify this and give me a direction to start working? Thanks!

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Let a, k , l , m e Z>1 and let a^k=1 (mod m) and a^l= 1 (mod m).
Let d=gcd(k,l)
Prove that a^d=1 (mod m).

I get already confused at the start: Is it true that k|phi(m) (Lagrange) but k can also be a multiple of the order of a (mod m) and then it can be the other way round.

Can anybody clarify this and give me a direction to start working? Thanks!
Hey tda120.

No. It is not necessary that $k|\phi(m)$. An easy counterexample is by taking $k=2\phi(m)$.

As for the question, use the fact that there exist integers $x$ and $y$ such that $kx+ly=d$.

I'd like to suggest that you check out our LaTeX forum and learn some basic LaTeX so that you will be able to post more readable questions.

#### tda120

##### New member
Thank you, I think this helps me a lot!
You're right; I should start using LaTeX, but I'm still a bit shy at using it...