Final answer:
To prove that n^12 ≡ 1 (mod 72), we consider the prime factorization of 72 and use Euler's totient function along with the Chinese Remainder Theorem. Since 72 is composed of the primes 2 and 3, we show n^12 ≡ 1 (mod 8) and n^12 ≡ 1 (mod 9), which is sufficient because n is coprime to 72.
Step-by-step explanation:
To prove the stronger statement that n^12 ⋅ 1 (mod 72), we must consider the properties of numbers and the modular arithmetic involved. Euler's theorem states that if n is coprime to 72, then n^{φ(72)} ⋅ 1 (mod 72), where φ(72) is Euler's totient function of 72. The totient φ(72) is 24 since 72 is factored into 23 × 32 and the totient function φ for powers of prime p is given by φ(pk) = pk − pk-1.
We can deduce the relationship n^12 ⋅ 1 (mod 72) by using the fact that 72 is a composite number whose prime factors are 2 and 3. By the Chinese Remainder Theorem, it suffices to show that n^12 ⋅ 1 (mod 8) and n^12 ⋅ 1 (mod 9). Since n is orthogonal to 72, n is odd and thus n^2 ⋅ 1 (mod 8) follows, and raising to the sixth power gives n^{12} ⋅ 1 (mod 8).
Similarly, for the mod 9 part, we know n^6 ⋅ 1 (mod 9) if n is coprime to 9 because φ(9) = 6. Raising both sides to the second power, we get n^{12} ⋅ 1 (mod 9) as well, completing the proof.