Question

[Hoàng Thị Minh Phương]

cmr \(t\left(x\right)=7^{2x+1}-48x-7⋮288\forall x\in N\)

Answer 1

[Akai Haruma]

Lời giải:

Vì \(7^3\equiv 1\pmod 9\) nên xét modulo $3$ cho $x$ :

+ Nếu \(x=3k\) :

\(\Rightarrow t(x)=7^{6k+1}-144k-7=7.7^{6k}-144k-7\equiv 7-144k-7\equiv 0\pmod 9\)

+ Nếu \(x=3k+1\):

\(\Rightarrow t(x)=7^{6k+3}-144k-55=7^3.7^{6k}-144k-55\equiv 7^3-55\equiv 0\pmod 9\)

+ Nếu \(x=3k+2\):

\(\Rightarrow t(x)=7^{6k+5}-144x-103=7^5.7^{6k}-144k-103\equiv 7^5-103\equiv 0\pmod 9\)

Từ 3 TH trên , suy ra \(t(x)\vdots 9\) $(1)$

Mặt khác:

\(t(x)=7(7^{2x}-1)-48x=7(7^x-1)(7^x+1)-48x\)

\( \bullet\) Nếu \(x\) chẵn, đặt $x=2t$ :

\(t(x)=7(7^t-1)(7^t+1)(7^x+1)-96t\)

+ $t$ lẻ:

\(\left\{\begin{matrix} 7^t-1\vdots 2\\ 7^x+1\vdots 2\\ 7^t+1\equiv (-1)^t+1\equiv 0\pmod 8\\ 96t\vdots 32\end{matrix}\right.\Rightarrow 7(7^t-1)(7^t+1)(7^x+1)-96t\vdots 32\)

\(\Rightarrow t(x)\vdots 32\)

+ $t$ chẵn:

\(\left\{\begin{matrix} 7^t-1\equiv (-1)^t-1\equiv 0\pmod 8\\ 7^x+1\vdots 2\\ 7^t+1\vdots 2\\ 96t\vdots 32\end{matrix}\right.\Rightarrow 7(7^t-1)(7^t+1)(7^x+1)-96t\vdots 32\)

\(\Rightarrow t(x)\vdots 32\)

\(\bullet \) Nếu \(x\) lẻ, đặt $x=2t+1$

Khi đó \(t=7(7^x-1)(7^x+1)-96t-48\)

Có \(\left\{\begin{matrix} 7^x+1\equiv (-1)^x+1= (-1)^{2t+1}+1\equiv 0\pmod 8\\ 7^x-1\vdots 2\\ 7^x-1\equiv (-1)^x-1=(-1)^{2t+1}-1\equiv -2\pmod 4\end{matrix}\right.\)

Do đó, \(7(7^x-1)(7^x+1)\) chia hết cho $16$ mà không chia hết cho $32$

Suy ra \(7(7^x-1)(7^x+1)=32k+16\Rightarrow t(x)=32k-96t-32\vdots 32\)

Từ 2TH trên, ta thu được \(t(x)\vdots 32(2)\)

Từ \((1),(2), UCLN(9,32)=1\Rightarrow t(x)\vdots (9.32=288)\) (đpcm)

\(\)