Consider the following statements: \(𝑆_1\): For any integer \(n>1, \:…

2019

Consider the following statements:

\(𝑆_1\): For any integer \(n>1, \: a^{\phi(n)} \equiv 1(mod \: n)\) for all \(a \in Z_n^*\) , where \(𝜙(𝑛)\) is Euler’s phi function.

\(𝑆_2\): If \(𝑝\) is prime, then \(a^p \equiv 1(mod \: p)\) for all \(a \in Z_p^*\).

Which one of the following is/are correct?

Answer: A. Only \(𝑆_1\)Answer: Only the first statement (S1) is correct. Explanation: First statement (S1): For any integer n>1 and any a coprime to n (a in Z_n^*), a^{ϕ(n)} ≡ 1…

  1. A.

    Only \(𝑆_1\)

  2. B.

     Only \(𝑆_2\)

  3. C.

     Both \(𝑆_1\) and \(𝑆_2\)

  4. D.

     Neither \(𝑆_1\) nor \(𝑆_2\)

Show answer & explanation

Correct answer: A

Answer: Only the first statement (S1) is correct.

Explanation:

  • First statement (S1): For any integer n>1 and any a coprime to n (a in Z_n^*), a^{ϕ(n)} ≡ 1 (mod n).

    This is Euler's theorem and is true under the stated hypothesis that a is in Z_n^* (i.e., gcd(a,n)=1).

  • Second statement (S2): If p is prime, then a^p ≡ 1 (mod p) for all a in Z_p^*.

    This is incorrect. The correct result for a prime p (Fermat's little theorem) is a^{p-1} ≡ 1 (mod p) for a in Z_p^*, which implies a^p ≡ a (mod p).

    Counterexample: take p = 5 and a = 2. Then 2^5 = 32 ≡ 2 (mod 5), not 1, so the claimed congruence a^p ≡ 1 fails.

Therefore only the first statement is true; the second statement is false as stated.

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