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Gia sư môn Toán » Gia sư Hỗ trợ giải toán THCS (6, 7, 8, 9), Số học

Cho p là số nguyên tố lớn hơn 3. CMR (p – 1)(p + 1) chia hết cho 24

[Bồi dưỡng HSG toán THCS] – Chuyên đề số nguyên tố. Đề bài: Cho p là số nguyên tố lớn hơn 3. CMR (p – 1)(p + 1) chia hết cho 24.

Giải:

TH1: p có dạng: p = 3k + 1 (k thuộc N*):

Ta có: (p – 1)(p + 1) = 3k.(3k + 2)

Vì p là số nguyên tố nên: k là số chẵn, k = 2n (Với n thuộc N*).

=> (p – 1).(p + 1) = 3.2n.(6n + 2) = 3.4.n.(3n + 1)

Nếu n là số lẻ thì 3n + 1 là số chẵn, ngược lại, n là số chẵn thì 3n + 1 là số lẻ nên suy ra: n.(3n + 1) chia hết cho 2

(p – 1)(p + 1) chia hết cho 3.4.2 = 24 (đpcm) (1)

TH2: p có dạng: p = 3k + 2 (k thuộc N*):

Ta có: (p – 1)(p + 1) = (3k + 1).(3k + 3) = 3.(3k + 1).(k + 1)

Vì p là số nguyên tố nên: k là số lẻ, k = 2n + 1 (Với n thuộc N*).

=> (p – 1).(p + 1) = 3.(6n + 4).(2n + 2) = 3.4.(3n + 2).(n + 1)

Nếu n là số lẻ thì 3n + 2 là số lẻ và n + 1 là số chẵn, ngược lại, n là số chẵn thì 3n + 2 là số chẵn và n + 1 là số lẻ nên suy ra: (3n + 2).(n + 1) chia hết cho 2.

=> (p – 1)(p + 1) chia hết cho 3.4.2 = 24 (đpcm) (2)

Từ (1) và (2) suy ra: (p – 1).(p + 1) chia hết cho 24 (đpcm).

Chúc các em học tập tốt 🙂

Mọi thông tin cần hỗ trợ đăng ký học tập bồi dưỡng Toán lớp 6, 7, 8, 9 vui lòng liên hệ Thầy Thích theo: 0919.281.916.

Thân ái!

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