- Nếu \(n=2k+1\Rightarrow9+2^{2k+1}=2.4^k+9\) chia 3 dư 2 ko thể là SCP
\(\Rightarrow n=2k\)
Đặt \(9+2^{2k}=x^2\Rightarrow x^2-\left(2^k\right)^2=9\)
\(\Rightarrow\left(x-2^k\right)\left(x+2^k\right)=1.9=3.3\)
Do \(0< x-2^k< x+2^k\) nên ta có:
\(\left\{{}\begin{matrix}x-2^k=1\\x+2^k=9\end{matrix}\right.\) \(\Rightarrow2.2^k=8\Rightarrow k=2\Rightarrow n=4\)