Perfect square involving the exponential law

Question

If $n$ is a natural number, and $2^{10} + 2^{13} + 2^n$ is a perfect square, what is the value of $n$?

I've attempted to factor out $2^{10}$ and got $2^{10}(1 + 2^3 + 2^{n-10})$. How can I move further?

Answer 1

$$(a+b)^2=a^2+2ab+b^2$$

$$2^{10}+2^{13}+2^n=(2^5)^2+2\times2^{12}+(2^7)^2=(2^5)^2+2\times2^{5+7}+(2^7)^2=(2^5+2^7)^2$$

Hence $n=14$

Answer 2

Complete the square:

$2^{10}+2^{13}+2^n=(2^5)^2+2\cdot2^5\cdot2^{n/2}+2^n=(2^5+2^{n/2})^2. $

$13=1+5+n/2$.

Can you take it from here?

Answer 3

We can factor out the $2^{10}$ first:

$$2^{10}(1+2^3+2^{n−10})$$

We can simplify $2^3 = 2\cdot{2}\cdot{2} = 8$. Also, because $2^{10}$ is a perfect square, we can remove it from the equation.

$$1+8+2^{n-10}=9+2^{n-10}$$

Now we can add a new variable, $m$, where $m$ can be any positive integer. We can rewrite our equation as:

$$9+2^{n-10}=m^2$$

We can now test values for $n-10$. If we do this, we'll find that $9 + 2^4 = 9+16=25=5^2$.

So $n-10=4$, and $n=14.$