(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(2n-1))+6^(3(2n-1))+9^(3(2n-1)) is always divisible by 3; hence there is no such square.
Edited on February 22, 2013, 3:27 am
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Posted by broll
on 2013-02-22 03:10:18 |
There is no solution.
