So $ 12 \mid (n - 1)(n + 1) $.

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• Solution: We are to find the smallest positive integer $ n $ such that:
• n^2 \equiv 1 \pmod{12}
• This implies $ n^2 - 1 \equiv 0 \pmod{12} $, or $ (n - 1)(n + 1) \equiv 0 \pmod{12} $.
• We test small values of $ n $:
• $ n = 1 $: $ 1^2 = 1 \equiv 1 \pmod{12} $ ✅
• Thus, the smallest such $ n $ is $ \boxed{1} $.

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