MHB Eulers phi function, orders, gcd

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The discussion centers on proving that if a^k ≡ 1 (mod m) and a^l ≡ 1 (mod m), then a^d ≡ 1 (mod m) where d = gcd(k, l). It is clarified that it is not necessary for k to divide φ(m), with a counterexample provided. Participants suggest using the relationship kx + ly = d to approach the proof. Additionally, there is a recommendation to utilize LaTeX for clearer communication of mathematical expressions. The conversation emphasizes understanding the properties of orders and gcd in modular arithmetic.
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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!
 
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tda120 said:
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.
 
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...
 
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