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
 A: 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.
A: If $n$ is odd then $2^n+3^n+6^n\equiv -1\pmod{3}$, so $2^n+3^n+6^n$ cannot be a square.
For similar reasons $\!\!\pmod{5}$, $n$ has to be a number of the form $4k+2$.
For similar reasons $\!\!\pmod{7}$, $n$ has to be a number of the form $12k\pm 2$, so $2^n+3^n+6^n$ is a multiple of $7$. $n=2$ gives the solution:
$$ 2^2+3^2+6^2 = 7^2. $$
Let we set $a_n=2^n+3^n+6^n$ and $\nu_7(m)=\max\{n\in\mathbb{N}:7^n\mid m\}$. 
Since $\frac{1}{4}+\frac{1}{9}+\frac{1}{36}=\frac{7}{18}$, if $n=12k-2$ then $\nu_7(a_n)\equiv 1\pmod{2}$, so $a_n$ cannot be a square and it is enough to study the case $n=12k+2$ and the sequence given by $b_n=a_{12n+2}$.
My bet now is on some variation of Zsigmondy's theorem. Numerical evidence gives that every $b_n$, for $n>0$, has a big prime divisor that appears with multiplicity one. If we manage to prove the last claim, we have that the previous solution ($n=2$) is the only one.
