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tda1201
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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!
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!