Simon asks: << Here is the problem. How to cover all N (natural numbers) with simple poynomials of the second degree even with some overlaps. We all know that it is possible to do it with a beatty sequence when 1/a + 1/b = 1, with a,b irrationals. [a*n] and [b*n] covers N with no overlab and no holes. if a = 1/2 + sqrt(5)/2 and b = 0.38196601125 then it works. Of course it is possible to do it with 2 or more ordinary arithmetical progressions. But how to do it with 2nd degree polynomials? Is it possible? For example (a tentative sieve that does not work), n^2 = 1, 4, 9, 16, 25, ... n^2+1 = 2, 5, 10, 17, 26, ... n^2+2 = 3, 6, 11, 18, 27, ... etc does cover many integers but I do not think it does work. It does not matter if there are some overlaps. of course, in the above example, n^2+k at n=1 will eventually reach any number but let's say : can it be done non-trivially? I am just wondering if the problem has a solution.
I'm not sure I understand just what the question is. Since for any polynomial P(x) over Z of degree > 1 we have density( { P(n) : n = 1,2,3,... } ) = 0, this clearly can't be done with finitely many polynomials. ---------------------------------------------------------------------- Let Sq denote {1,4,9,16,...}. But if say for each k in Z, we let P_k(x) = x^2 + kx , then { P_k(n) : n,k = 1,2,3,... } = Sq + { kn : k,n = 1,2,3,... } then since every positive integer is a square plus a number of the form kn, this is certainly Z (with abundant overlap). So, I'm not sure if I'm understanding the question. --Dan