Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it. 
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.

If $0\le n\le 7$, then there are no solutions. Let $n\ge 8$.
$$2^8+2^{11}+2^n=\left(2^4\right)^2\left(9+2^{n-8}\right)$$
is a square if and only if $9+2^{n-8}=m^2$ for some $m>3$, i.e. $2^{n-8}=(m+3)(m-3)$, so $m+3=2^k$ and $m-3=2^l$ for some $k>l\ge 0$. If $k\ge 4$, then $$6=2^k-2^l\ge 2^k-2^{k-1}\ge 8>6$$ contradiction, so $k\in\{1,2,3\}$, which only gives $k=3$, so $m=5$, $n=12$.
