If there is such a square, it must be worth 1 mod 2, and 1 mod 3. Solutions to 4^{n}+6^{n}+9^{n }are cyclic mod 5 {4,3} but no square is worth 3 mod 5, and cyclic mod 7 {5,0,1,0,5,3}, but no square is worth 3 or 5 mod 7. The upshot of this is that only values of n that are divisible by 3 could potentially qualify as squares. No doubt this process could be continued...
Edited on February 22, 2013, 3:25 am

Posted by broll
on 20130222 02:47:36 