Question

If a, b are integers and p is a prime number, then which of the following statements is/are false?
(I) If \((a,p) = (b,p) = 1\) and \(a^k \equiv b^k \pmod{p}\) (\(k \in \mathbb{N}\)), then \(a \equiv b \pmod{p}\).
(II) If \((a,p) = 1\) and \(ax \equiv 1 \pmod{p}\), then \(a \equiv a^{p-2} \pmod{p}\).

• A. Only (I)
• B. Only (II)
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

Try a root-of-unity counterexample for (I), and remember Fermat's theorem gives x, not a, equal to a to the power (p-2).

Answer 1

The Correct Option is C

Step 1: Test statement (I) with a counterexample. Take \(p = 7\), \(a = 2\), \(b = 4\); both are coprime to \(7\). The order of \(2\) modulo \(7\) is \(3\), since \(2^3 = 8 \equiv 1 \pmod 7\), and \(4 = 2^2\) also satisfies \(4^3 = (2^2)^3 = 2^6 \equiv 1 \pmod 7\). So at \(k = 3\), \(a^k \equiv 1 \equiv b^k \pmod 7\), yet \(a = 2\) and \(b = 4\) are not congruent modulo \(7\). This contradicts the claim, so statement (I) is false. The reason is that \(a^k \equiv b^k \pmod p\) only forces \((ab^{-1})^k \equiv 1 \pmod p\), so \(ab^{-1}\) is a \(k\)-th root of unity, and this root need not equal \(1\) whenever \(\gcd(k, p-1) > 1\).

Step 2: Test statement (II) as stated. By Fermat's Little Theorem, since \((a,p) = 1\), \(a^{p-1} \equiv 1 \pmod p\), so \(a \cdot a^{p-2} \equiv 1 \pmod p\). This shows \(a^{p-2}\) is the multiplicative inverse of \(a\) modulo \(p\), that is, the solution \(x\) of \(ax \equiv 1 \pmod p\) satisfies \(x \equiv a^{p-2} \pmod p\). The statement instead claims \(a \equiv a^{p-2} \pmod p\), equating \(a\) itself, not \(x\), with \(a^{p-2}\), which would require \(a^{p-3} \equiv 1 \pmod p\).

Step 3: Check with a concrete case: \(p = 7\), \(a = 3\). Here \((3,7)=1\) and \(3x \equiv 1 \pmod 7\) has solution \(x = 5\), since \(3 \times 5 = 15 \equiv 1 \pmod 7\). Now compute \(a^{p-2} = 3^5 = 243 \equiv 5 \pmod 7\), since \(243 = 34 \times 7 + 5\). The claim asserts \(a \equiv a^{p-2}\), that is \(3 \equiv 5 \pmod 7\), which is false. So statement (II), exactly as written, is false.

Step 4: Both statements (I) and (II) are false.

\[\boxed{\text{Both (I) and (II)}}\]