\(2^n+2^8+2^{11}=m^2\Leftrightarrow2^n+2^8.\left(2^3+1\right)=m^2\)
\(\Leftrightarrow2^n+2^8.3^2=m^2\)
\(\Leftrightarrow2^n+\left(2^4.3\right)^2=m^2\)
\(\Leftrightarrow2^n+48^2=m^2\)
\(\Leftrightarrow2^n=\left(m-48\right)\left(m+48\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m-48=2^a\\m+48=2^b\end{matrix}\right.\) với \(\left\{{}\begin{matrix}a;b\in N\\b>a\end{matrix}\right.\)
\(\Rightarrow2^b-2^a=96\)
\(\Rightarrow2^a\left(2^{b-a}-1\right)=96=2^5.3\)
Do \(2^{b-a}-1\) luôn lẻ nên \(2^{b-a}-1\) là ước lẻ của 96
\(\Rightarrow\left[{}\begin{matrix}2^{b-a}-1=1\\2^{b-a}-1=3\end{matrix}\right.\)
- Nếu \(2^{b-a}-1=1\Rightarrow2^a=96\Rightarrow\) ko tồn tại a thỏa mãn
- Nếu \(2^{b-a}-1=3\Rightarrow\left\{{}\begin{matrix}2^a=2^5\\b-a=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=5\\b=7\end{matrix}\right.\)
\(\Rightarrow n=12\)
