If there is such a square, it must be worth 1 mod 2, and 1 mod 3. Solutions to 4n+6n+9n 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
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Posted by broll
on 2013-02-22 02:47:36 |
There is probably no solution
