2016 is absolutely awesome 12

I assumed that since n was the largest positive integer, it would be greater than 2016.

$2^{2016}+2^{6012}+2^n$ can be written as $2^{2016}.(2^{n-2016} + 2^{3996}+1)$

Now $2^{2016}$ is already a square number, so for the second bracket we can use the following expansion:

$(2^{m}+1)^{2}= 2^{2m}+2^{m}+1$

Therefore:

$2m=n-2016$

$m+1=3996$

Since doing the other substitution, i.e:

$2m=3996$

$m+1=n-2016$

yields a smaller n, i.e: n= 4015.

Therefore,On solving n comes out as 10006

We can rewrite the equation as the following:

$(2^{1008}+2^{3006})^2=2^{2016}+2^{3006}+2^n$ in which case $n=4015$

Or another possibility is:

$(2^{1008}+2^{\frac{n}{2}})^2=2^{2016}+2^{6012}+2^{n}$ . In this case, $n=10006$

Clearly the second case yields a way larger value of $n$

I get it that $n= 10006$ satisfy the expression to be a perfect square. What I dont get is proving that $10006$ is the largest solution.

Compare the given expression to $\large (2^m)^2$ + $\large (2^p)^2$ + $\large 2(2^m)(2^p)$ = $\large (2^m + 2^p)^2$ , where $\large 2p = n$ . Two cases arise, one when $\large (2^{6012})$ is the $\large 2(2^m)(2^n)$ term and second when its a squared term.

You shall notice that the former case gives a larger value for n.