Rich Schroeppel wrote:

When N = 6K+1, (N^2+N+1)^2 | 3 * [(N+1)^(N+1) - N^N]; ... When N = 6K+4, (N^2+N+1)^2 | 3 * [(N+1)^(N+1) + N^N]. ... True or False?

I haven't checked the details, but after a bit of scribbling I'm fairly sure the answer is ... ... ... ... ... ... ... True. The trick (or at least, the trick I used; there may well be a better way to approach it) is to say things like (N+1)^2 = N + (N^2+N+1) (N+1)^(6K+4) = (N + (N^2+N+1))^(3K+2) = N^(3K+2) + N(3K+2)(N^2+N+1) mod (N^2+N+1)^2 (using the binomial theorem),and now you can use N=6K+4 to advantage. Similarly, N^3 = 1 + (N-1)(N^2+N+1), and again you can raise both sides to whatever power is convenient and apply the binomial theorem. As I say, I haven't actually done all the necessary algebra, but (1) it looks like it probably works out and (2) a 30-second numerical check for the first 100000 values of N says it's good, so I bet the details check out. -- g