What I wrote is nonsense, but what does make sense is the following: Let Q_k(x) = x^2 + kx for all k in Z. Then: S+ = {Q_k(p) : p prime, k = 1,2,3,... } contains for each prime p all multiples of p greater than p^2, S- = {Q(-k)(p) : p prime, k = 1,2,3,...} contains for each prime p all multiples of p less than p^2, and S0 = {Q_0(n) : n = 1,2,3,...} contains all squares. So S- u S0 u S+ = Z. So I'm guessing that Simon wants a family of polynomials { P_k(x) : k = 1,2,3,... } with all coefficients positive integers with { P_k(n) : k = 1,2,3,... and n = 1,2,3,...} = Z+ exactly. -------------------------------------------------------------------------------- Of course, as long as we can use countably many polynomials, for each k we can always design P_k(x) so that P_k(1) = [the least positive integer not contained in the union of {P_j(n) : n = 1,2,3,...} for 1 <= j < k.] But I suppose this would be considered too trivial. --Dan ------------------------------------------------------------------ I wrote: << Let Sq denote {1,4,9,16,...}. But if say for each k in Z, we let P_k(x) = x^2 + kx , then . . . . . . . . .

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