Ed, et al:
Define the n-complement of a number j as the number k, such that corresponding digits of j and k always add to n.
In other words, j+k = n*(10^a-1)/9, for some A, and all digits of J and K are <= N, < 10, >= 0, and > N-10.
( 225 , 441 ) are 6-complements, and both are square numbers ( 39866596 , 82355625 ) are 11-complements, and both are square numbers 2. What is the next set of n-complementary square numbers?
This is a lot like representing X=n*(10^a-1)/9 as the sum of two squares. (In your case, you also want all digits to be in-bounds.) I think the two-square-sum problem has been done to death. If you can factor X, it's easy to determine whether it has such representations; does anyone know how to generate them? -- Don Reble djr@nk.ca