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 $4=2^2$.
 A: $2^8 + 2^{11} + 2^n = 2^8(1 + 8 + 2^{n-8})=2^8(9 + 2^{n-8})$
Therefore, $9 + 2^{n-8}$ has to be a perfect square. 
Clearly, $9 + 16 = 25$ is a perfect square. 
So, $2^{n-8} = 2^4$ giving, $$n = 12$$
A: 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 $n=12$. 
A: See we need an integer power of $2$ so difference between a square and know indices should be even now $2$ follows a particular order of last digits which is $2,4,8,6...$ so now we need numbers which are even which will give difference as these above numbwrs .Now we know no perfect square ends in $2,8$ so now we are left with $4,6$ but(last digit) $4-4=0$ no power of $2$ gives $0$ as last digit so now we need to check for $6$ now when we plug $60^2,70^2..$ we get our last digit as $6$ ie eg $3600-2304=...6$ so now checking these cases we get $n=80$ so $80.80=6400$ $2^8+2^{11}=2304$ so $6400-2304=4096$ which is $2^{12}$ thus $n=12$
A: Hint: $(2^a + 2^b)^2 = 2^{2a} + 2^{2b} + 2^{a+b+1}$.  
A: Here is a very late answer since I just saw the problem:
By brute force, we may check that $n=12$ is the smallest possible integer such that $2^8+2^{11}+2^n$ is a perfect square. We also claim that this is the only integer.
To see why the above is true, let $2^8+2^{11}+2^n=2^8(1+2^3+2^k)=2^8(9+2^k), k \ge 4$. Now, we only need to find all integers $k \ge 4 $ such that $9+2^k$ is a perfect square. Let $m^2=9+2^k, m \ \in \mathbb{Z^+}$. Hence we have $2^k=(m-3)(m+3)$. Clearly, $m$ must be odd; otherwise, both $m-3$ and $m+3$ would be odd, contradiction. Hence, $2\mid m-3$ and $2 \mid m+3$. But we also claim it is impossible that both $m-3$ and $m+3$ divides $4$; otherwise, $4  \mid 2m \Rightarrow 2 \mid m$, which is a contradiction for $m$ odd. Since $m-3 < m+3$, we conclude that, for any integer $k \ge 4$, we must have $m-3=2$ and $m+3=2^{k-1}$. But the former already implies $m=5$, so we must have $k-1=3 \Rightarrow k=4$ as our only solution, which is precisely what we claimed earlier.
A: 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 $m=5$, $n=12$.
A: You can write
$$2^8+2^{11}+2^n=(2^4)^2+2.2^4.2^6+2^n$$
now, note that if $n=12$, follows that
$$(2^4)^2+2.2^4.2^6+2^{12}=(2^4)^2+2.2^4.2^6+(2^6)^2=(2^4+2^6)^2$$
Thus, $n=12$ to solve the problem.
A: Solution :
$2^n+2^8+2^(11)=m^2$
$$2^n=m^2-2^8-2^(11)$$
$$2^n=m^2-2^8(1+2^3)$$
$$2^n=m^2-2^8*9$$
$$2^n=m^2-(2^4*3)^2$$
$$2^n=m^2-48^2$$
$$2^n=(m-48)(m+48)$$
$$2^(n-k+k)= (m-48)(m+48)$$ 
$$2^(n-k)*2^k= (m-48)(m+48)$$
So
(1) $$2^(n-k)= (m-48)$$ and
(2) $$2^k= (m+48)$$
(2)-(1)   $$2^k-2^(n-k)=(m+48)-(m-48)$$
              $$2^k-2^(n-k)=96$$
              $$2^k-2^(n-k)=2^5*3$$
              $$2^(n-k)*(2^(2k-n)-1)= 2^5*3$$
              $$2^(n-k)=2^5$$  and  $$2^(2k-n)-1=3$$
              $$n-k=5$$            and   $$2^(2k-n)=2^2$$
              $$n-k=5$$            and   $$2k-n=2$$
              $${n=12,k=7}$$
$$n=12$$
