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