If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$. If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$.

My first shot would be to assume the perfect square is $2^{2018}$, but how would I prove that?  Even if it is, what is $n$?  All help is appreciated.

$(2^x+2^y)^2=2^{2x}+2^{x+y+1}+2^{2y}$ so we can take $y=1007$ and $x=1009$ to conclude $2^{2018}+2^{2017}+2^{2014}$ is a square.

how to obtain all $n$:
$9(2^{2014})=2^{2017}+2^{2014}=k^2-2^n$.
Notice that we must have $k^2\equiv 1 \bmod 3$ implying $2^n\equiv 1\bmod 3$.
So $n=2m$.
From here we have $2^{2014}\times 9= (k+2^m)(k-2^m)=z(z+2^{m+1})$
there are now two cases, case one is when $z=9(2^j)$ and case two is when $z=2^j$
the first case is impossible because the number $9$ in binary is $1001$ and adding a power of two clearly wont make it a power of $2$. The other case is clearly only solved by taking $z=2^{1007}$ and taking $m=1009$ thus getting $z(z+2^{m+1})=2̣^{1007}(2^{1007}+2^{1010})=9(2^{2014})$.
So we need $n=2(1009)=2018$
