1 reply to this topic

[dungxibo123]

chung minh rang neu $n=\frac{(a^{2p}-1)}{a^2-1}$, trong do $a$ ;a so nguyen, $a>1$ va $p$ ;a so nguyen to $>2$, $a(a^2-1)$ khong chia het cho $p$. Thi $n$ la so gia nguyen to co so $a$.
thanh that xin loi moi nguoi vi unikey cua may em bi hong

Edited by dungxibo123, 21-03-2017 - 15:37.

[yeutoan2001]

Bài toán qui về chứng minh:

   $a^{n-1}-1\vdots n$

Thật vậy ta có 

  $a^{2(p-1)}-1\vdots p$

MÀ ($a(a^2-1),p$)=1 nên

Do n lẻ nên n-1 chia hết cho 2 

  Suy ra n-1=$\frac{a^{2p}-a^2}{a^2-1}\vdots 2p$

=> 

        $a^{n-1}-1\vdots a^{2p}-1\vdots \frac{a^{2p}-1}{a^2-1}$

Edited by yeutoan2001, 21-03-2017 - 17:20.