# When does a polynomial $f(x)$ generate an arithmetic sequence for consecutive values of integer $x$?

Given polynomial $$f(x) = a_0 + a_1x + \dots + a_n x^n, a_n \ne 0, n \ge 2, k \in (0,n), a_k \in \{0, 1, 2, \dots\}$$, under what conditions is $$f(r), f(r+1), \dots f(r+n)$$ an arithmetic sequence for integer $$r$$?

When does $$f(x)$$ not generate an $$(n+1)$$ term arithmetic sequence? (Update: Never - based on responses to this question).

Updated question: When does $$f(x)$$ generate an $$n$$ term arithmetic sequence? Given n values, we can fit an $$n$$-degree polynomial. The question is given the polynomial, when does it generate an $$n$$-term arithmetic sequence for $$n$$ consecutive values of $$x$$

Note that $$a_k$$ are fixed in this problem.

• I've never studied discrete differences, so I am giving you a blind guess that may easily be wrong. Since polynomials are continuous functions, if $g(x) = f(x+1) - f(x)$, then it seems to me that you want $g'(x) = 0.$ – user2661923 Sep 30 '20 at 14:02
• Re Rivers McForge's answer, looks like I got lucky. – user2661923 Sep 30 '20 at 14:06

In one direction, this obviously happens for every $$r$$ when $$f(x)$$ is linear. In the other direction, suppose $$f(x)$$ is not linear, then we can write the $$n+1$$ values $$f(r), f(r+1), ..., f(r+n)$$ as $$a + bx$$ for $$x = r, r+1, ..., r+n$$. Then the polynomial $$g(x) = f(x) - (a + bx)$$ is a nonzero polynomial of degree $$n$$ but has $$n+1$$ zeroes, which is impossible by the Fundamental Theorem of Algebra. So, such an arithmetic sequence exists if and only if $$f(x)$$ is linear.
• You don’t need to invoke the fundamental theorem of algebra - it is an easy fact that a degree $n$ nonzero polynomial has at most $n$ zeros. – Joppy Sep 30 '20 at 14:25
• Ok. What happens if we consider a $n$-term arithmetic sequence instead of $(n+1)$-term sequence? – vvg Sep 30 '20 at 14:37
• @vvgiri The exact same argument shows that it's linear. You'd need less than $n$ terms for a non-constant degree $n$ polynomial. – Rivers McForge Sep 30 '20 at 15:13
If the sequence has $$n$$ terms, one can fit a $$n + 1$$ or higher degree polinomial that agrees on that points (and has arbitrary values at other, selected points).