On the Degree of Univariate Polynomials Over the Integers

摘要

study the following problem raised by von zur Gathen and Roche: What is the minimal degree of a nonconstant polynomial f:0n0m ? Clearly, when m=n the function f(x)=x has degree 1. We prove that when m=n?1 (i.e. the point n is not in the range), it must be the case that deg(f)=n?o(n). This shows an interesting threshold phenomenon. In fact, the same bound on the degree holds even when the image of the polynomial is any (strict) subset of 0n . Going back to the case m=n, as we noted the function f(x)=x is possible, however, we show that if one excludes all degree 1 polynomials then it must be the case that deg(f)=n?o(n). Furthermore, the same conclusion holds even if m=O(n1475?). In other words, there are no polynomials of intermediate degrees that map 0n to 0m . Moreover, we give a meaningful answer when m is a large polynomial, or even exponential, in n. Consider the case m=1d!2en?dd . f can of course be a degree d?1 polynomial, e.g., f(x)=xd?1 or even f(x)=d?1x?n2 (whose range is bounded by n2d?1 ). We show that when d2n15 , either deg(f)d?1 or f must satisfy deg(f)n3?O(dlogn) . Stated differently, if we remove the `trivial' cases where f is of degree at most d?1, the next example must have a very high degree. So, again, no polynomial of intermediate degree mapping 0n to 01d!2en?dd exists. We complement these results by showing that for every d=o(nlogn) there exists a polynomial f:0n0O(nd+05) of degree n3?O(dlogn)deg(f)n?O(dlog(n)) . Our work considerably extends the results of von zur Gathen and Roche that studied the case m=1.