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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.

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    $\begingroup$ 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 $g'(x) = 0.$ $\endgroup$ – user2661923 Sep 30 '20 at 14:02
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    $\begingroup$ Re Rivers McForge's answer, looks like I got lucky. $\endgroup$ – user2661923 Sep 30 '20 at 14:06
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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.

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    $\begingroup$ 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. $\endgroup$ – Joppy Sep 30 '20 at 14:25
  • $\begingroup$ Ok. What happens if we consider a $n$-term arithmetic sequence instead of $(n+1)$-term sequence? $\endgroup$ – vvg Sep 30 '20 at 14:37
  • $\begingroup$ @vvgiri The exact same argument shows that it's linear. You'd need less than $n$ terms for a non-constant degree $n$ polynomial. $\endgroup$ – Rivers McForge Sep 30 '20 at 15:13
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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).

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