Timeline for Four polynomials representing all integers modulo m

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Nov 16, 2010 at 19:00	comment	added	David E Speyer		@Victor Miller: Asking for only a finite set of such m will require additional conditions. For example, take the polynomials $k^2$ and $-k^2$. For any prime which is $3 \mod 4$, the union of their images is everything.
Nov 12, 2010 at 23:41	comment	added	Felipe Voloch		Victor, if $y \equiv f(k) \mod m$ then $y \equiv f(k) \mod p$ if $p\|m$. I am not using CRT. But for arbitrary polynomials, the complement of their image is not the projection of a curve in one of its coordinates, so yes, your generalization will be harder.
Nov 12, 2010 at 21:04	comment	added	Victor Miller		Added: in this specific case, since for any $m>2$ there's always a $c$ which is a quadratic non-residue mod all the primes dividing $m$ a CR type argument can be made to work. I don't see how to do it in the general case in my last comment.
Nov 12, 2010 at 21:02	comment	added	Victor Miller		Good start. However, since the set that we're looking at is the union of images of polynomials, a Chinese remainder theorem type argument doesn't work. An interesting generalization of the original questions is given a finite set of polynomials in $\mathbb{Z}[x]$ (say of degree $>1$) is there only a finite set of $m$ such that every point $\bmod m$ is the image of the value of one of the polynomials?
Nov 11, 2010 at 19:05	vote	accept	Martin Erickson
Nov 11, 2010 at 15:15	history	answered	Felipe Voloch	CC BY-SA 2.5