Perfect Square Problem

We can use the identity $(a+b)^2=a^2+2ab+b^2$ .Let's try to put the expression into this form: $2^8+2^{11}+2^n\\=(2^4)^2+2(2^4)(2^6)+2^n$ We can see that in order to be a perfect square trinomial of the above mentioned form, $2^n$ must be $(2^6)^2=2^{12}$ .As bases are same so the exponent,so $\boxed{n=12}$

$2^{11}+2^{8}+2^{n}$

$= 2^{8}(2^{3}+1+2^{(n-8)})$

$= (2^{4} \times 2^{4})(8+1+2^{(n-8)})$

$= (2^{4} \times 2^{4}) (9+2^{(n-8)})$

Since, $(2^{4} \times 2^{4})$ is a Perfect Square,

$(9+2^{(n-8)})$ must also be a Perfect Square.

Let $(n-8) =$ m,

Then, $9+2^{m} =$ Perfect Square

When m $= 4$ ,

$9+2^{4} = 9+16 = 25$

So, $n-8 = 4$

$n = 4+8 = \boxed{12}$

If $2^{11}+2^8+2^n=m^2$ , for $m$ an integer.

$2^8\cdot(2^3+1+2^{n-8})=m^2$ .

From the prime factorization rule, it follows that;

$2^3+1+2^{n-8}=k^2$ for some integer $k$ .

Now $2^{n-8}=k^2-9$ , thus $2^{n-8}=(k-3)(k+3)$

But since $2^{n-8}$ is a power of $2$ , so are $(k-3)$ and $(k+3)$

Say $k-3=2^t$ and $k+3=2^s$ , for natural numbers $s,t$ then we have;

$6=2^s-2^t\implies2^t(2^{s-t}-1)=2\cdot3$

Thus $t=1, s=3$ and plugging these values into the equations we obtain $k=5$ and $n=12$

$2^{11}+2^8 +2^n = 2^8 (1+2^3) + 2^n = 9\times\ 2^8 +2^n$ Now we need to factorise again by using using a well known pythagoraen triple (namely the 3,4,5 triple) : $2^8(9+16) = 2^8(9+2^4) = (9\times\ 2^8) +2^{12}\therefore\boxed{n = 12}$ Substituting n =12 makes the expression equal to $6400 = 80^{2}$