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)
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