Đặt \(2^4+2^7+2^n=a^2\) (a \(\in\) N)
\(\iff\) \(\left(2^4+2^7\right)+2^n=a^2\)
\(\iff\)\(2^4.\left(1+2^3\right)+2^n=a^2\)
\(\iff\)\(2^4.3^2+2^n=a^2\)
\(\iff\)\(\left(2^2.3\right)^2+2^n=a^2\)
\(\iff\) \(12^2+2^n=a^2\)
\(\iff\)\(2^n=a^2-12^2\)
\(\iff\)\(2^n=\left(a-12\right).\left(a+12\right)\)
Đặt \(a-12=2^q\left(2\right)\) \(;a+12=2^p\left(1\right)\)
Gỉa sử :p>q ,p,q \(\in\) N
Lấy (1)-(2) vế với vế ta được \(24=2^p-2^q\)
\(2^3.3=2^q.\left(2^{p-q}-1\right)\)
\(\implies\) \(\hept{\begin{cases}2^3=2^q\\3=2^{p-q}-1\end{cases}}\) \(\implies\) \(\hept{\begin{cases}q=3\\2^2=2^{p-q}\end{cases}}\) \(\implies\) \(\hept{\begin{cases}q=3\\p-q=2\end{cases}}\) \(\implies\)\(\hept{\begin{cases}q=3\\p=5\end{cases}}\)
\(\implies\) \(n=p+q=3+5=8\)
Với n=8 thì \(2^4+2^7+2^n=2^4+2^7+2^8=16+128+256=400=20^2\) là số chính phương thỏa mãn ycbt
Vậy n=8