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 $4=2^2$.

Note the fact that $2^8+2^{11}=48^2$
This implies that we are trying to find values of $n$ where $2^n=(x-48)(x+48)$. 
Thus, we must find $k,l$ where $2^k-2^l=96$(where $x+48=2^k$, $x-48=2^l$)
Note the fact that $k \ge 7$. 
This implies that $2^k$ is divisible by $32$, which implies that $2^l$ is also divisible by $32$.  
Also, notice that if $k \ge 9$, when $2^k-2^l \ge 256$.
This implies that $k=7$or $k=8$. 
Note the fact that $k=8$ does not have an integer solution, thus $k=7$. 
Thus $x=80$, and thus $n=12$. 
