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2023-11-25 10:48:17
Vì p là số nguyên tố lớn hơn 3 → p không chia hết cho 3 → p có dạng 3k + 1 hoặc 3k + 2.
Với p = 3k + 1
→ (p - 1)(p + 1) = 3k(3k + 2) ⋮ 3
Với p = 3k + 2
→ (p - 1)(p + 1) = 3(3k - 1)(k + 1) ⋮ 3
⇒ (p - 1)(p + 1) luôn chia hết cho 3 (1)
Mà p là số nguyên tố → p là số lẻ → (p - 1)(p + 1) là tích hai số chẵn liên tiếp.
⇒ (p - 1)(p + 1) chia hết cho 8 (2)
Mà (3, 8) = 1 (3)
Từ (1), (2) và (3) → (p - 1)(p + 1) ⋮ 24
Tương tự CM: (q - 1)(q + 1) ⋮ 24
⇔ (p - 1)(p + 1) - (q - 1)(q + 1) ⋮ 24
⇔ p2 - 1 - q2 + 1 ⋮ 24
⇔ p2 - q2 ⋮ 24 (đpcm)