If $n,k\in\mathbb N$, solve $2^8+2^{11}+2^n=k^2$. If $n,k\in\mathbb N$, solve $$2^8+2^{11}+2^n=k^2$$
It's hard for me to find an idea. Some help would be great. Thanks.
 A: HINT:
We have $\displaystyle2^8+2^{11}=2304$
Now, $2^8+2^{11}+2^0=2304+1\ne k^2$
$\displaystyle 2^8+2^{11}+2^1=2304+2\equiv2\pmod8$, but $a^2\equiv0,1,4\pmod8$
So,  $n\ge2$ let $n=m+2$ where $m\ge0$
$\displaystyle 2^8+2^{11}+2^{m+2}=4(576+2^m)\implies 576+2^m$ must be perfect square
Like either method $m\ge2$ let $m=r+2$ where $r\ge0$
$\displaystyle576+2^{r+2}=4(144+2^r)$ 
Follow this step
One Observation :
$$(2^4)^2+2\cdot2^4\cdot2^6+(2^6)^2=(2^4+2^6)^2$$
A: First assume $n\ge 8$ and write $n=m+8$, we get:
$$(2^4)^2(1+2^3+2^m)=k^2$$
This is possible only if $1+2^3+2^m=9+2^m$ is a perfect square. If it is, then we can write it as $(3+p)^2$ (since it is obviously greater than $3$), and thus:
$$9+6p+p^2=9+2^m$$
$$\Leftrightarrow p(6+p)=2^m$$
In particular, $p=2^r$ and $6+p=2^{m-r}$ for some $1\le r\le m$ (notice that $r=0$ is impossible). From this we get $6+p=6+2^r=2(3+2^{r-1})=2^{m-r}$, and the only solution to this equation is $r=1$, $m=4$. Thus the only solution for $n\ge8$ is $n=12$, which gives us:
$$2^8+2^{11}+2^{12}=80^2$$
Now for $0\le n<8$, we have:
$$2^n(2^{8-n}+2^{11-n}+1)=k^2$$
Since the number in the brackets cannot be a multiple of $2$, we observe that $n$ must be even. This leaves us with the condition that $2^{8-n}+2^{11-n}+1$ is a square. That number is exactly:
$$2^{8-n}3^2+1$$
For $n=6$ we have $2^23^2+1=37$ which is not a square.
For $n=4$ we have $2^43^2+1=145$ which is not a square.
For $n=2$ we have $2^63^2+1=577$ which is again not a square.
Finally for $n=0$ we have $2^83^2+1=1153$ which is not a square either.
Thus the only solution is $n=12$.
A: This is justa a partial solution.
we assume first that $n>8$ and we see $2^8+2^{11}+2^n=k^2$ as
$2^8(1+2^3+2^{n-8})=k^2$ we see $2^8$ is a square...
so the problem reduces to see when would $1+2^3+2^{n-8}$ be a square..
As we have assumed $n>8$ we would have $1+2^3+2^{n-8}$ to be an odd integer.
Now, any odd perfect square is congruent to $1$ modulo $8$
we see that $1+2^3$ is congruent to $1$ modulo $8$
so the problem is to make $2^{n-8}$ as $0$ modulo $8$
i.e., to make $n-8=3k$ 
least possible case would be $k=1$ i.e., $n=12$
I tried to repeat this and let $k=2$ but then It does not work.
being congruent to $1$ mod $8$ is only necessary condition but not sufficient so I can not say this works for all $k\geq 2$.
trying to solve this completely for a long time with no success. so i thought of posting what i have done and expecting for some suggestions.
