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

. . .

. . .

. . .

>>