(In reply to
There is probably no solution by broll)
No doubt this process could be continued... as far as 19 (=4+6+9) gives: {0,0,2,0,0,14,0,0,3,0,0,2,0,0,14,0,0,3}. The significance is that every solution, where n is divisible by 3, is either 2, 3, or 14, mod 19, whereas none of these can be squares mod 19.
But the 'potentially qualifying power' in 4^(3(2n1))+6^(3(2n1))+9^(3(2n1)) is always divisible by 3; hence there is no such square.
Edited on February 22, 2013, 3:27 am

Posted by broll
on 20130222 03:10:18 