When is $2^n +3^n + 6^n$ a perfect square? Find all $n$ for which $2^n+3^n+6^n$ is a perfect square.
 I  do not have a specific idea how to solve this one

If $2^n+3^n+6^n=x^2$ for some positive integer $x$, then we know from a comment by user236182 that $n=2k$ for some positive integer $k$.  Hence,
$$\left(x-2^k\right)\left(x+2^k\right)=3^{2k}\left(1+2^{2k}\right)\,.$$
Since $x$ is obviously odd, $\gcd\left(x-2^k,x+2^k\right)=1$, so $3^{2k}$ divides either $x-2^k$ or $x+2^k$.  However, $3^{2k}>1+2^{2k}$ implies that $x+2^k=m\cdot 3^{2k}$ for some positive integer $m$.  If $m\geq 2$, then $$2x=\left(x+2^k\right)+\left(x-2^k\right)>x+2^k=m\cdot 3^{2k}\geq 2\cdot 3^{2k}$$ leads to $x>3^{2k}$.  Therefore, $$9^{2k}<x^2=2^{2k}+3^{2k}+6^{2k}<9^{2k}\,,$$
which is absurd.  Hence, $m=1$.  That is, $x+2^k=3^{2k}$ and $x-2^k=1+2^{2k}$.  Therefore, $$3^{2k}=x+2^k=\left(1+2^k+2^{2k}\right)+2^{k}=1+2^{k+1}+2^{2k}\,.$$  If $k>1$, then $1+2^{k+1}+2^{2k}<3^{2k}$, which is a contradiction.  Hence, $k=1$, or $n=2$, is the only possible solution.  As $n=2$ yields $x=7$, we are done.
