Question

Generalization of Euler’s theorem). Prove that a^m≡a^(m−φ(m)) (mod m).

Answer 1

To prove the congruence a^m ≡ a^(m-φ(m)) (mod m), use Euler's theorem.

Step-by-step explanation:

To prove that a^m ≡ a^(m-φ(m)) (mod m), we need to use Euler's theorem.

1. If a and m are coprime, then Euler's theorem states that a^φ(m) ≡ 1 (mod m).
2. Using this, we can write a as a^φ(m) * a^(m-φ(m)).
3. By taking both sides of the congruence raised to the power of m, we get (a^φ(m))^m * (a^(m-φ(m)))^m ≡ 1^m (mod m).
4. Since a^φ(m) ≡ 1 (mod m), the first term becomes 1.
5. Therefore, we are left with a^(m-φ(m)) ≡ 1 (mod m) which proves the given congruence.