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I was working with a student on the following problem (source unknown)

Let $f(n)$ be the minimum degree of a monic polynomial $p$ such that for all integers $m$, $p(m)$ is a multiple of $n$. Evaluate $f(10^{10})$.

Clearly, $f(10^{10})\leq45$ as $$p(x)=\prod\limits_{k=0}^{44}(x+k)$$satisfies the requirements, but we are struggling to show that this is optimal (or if not, how we can do better).

Rushabh Mehta
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    Interesting. What about smaller values? Is the answer always given by the least integer $k$ with $n,|,k!$ ? – lulu Mar 21 '22 at 19:06
  • @lulu That's my theory, but I've failed to prove it. – Rushabh Mehta Mar 21 '22 at 19:06
  • Well, for small numbers it shouldn't be hard to check, no? Try $6$ for instance. – lulu Mar 21 '22 at 19:07
  • @lulu It's true for $6$. I just brute forced all of the possibilities. – Rushabh Mehta Mar 21 '22 at 19:08
  • here is a (closely) related question with some results and some links (which I did not study). – lulu Mar 21 '22 at 19:24
  • @lulu From that link, it seems as though it is believed that our theory is correct, although the proof is not known. I don't get how this question is meant to be solved by high schoolers in that case, lol. – Rushabh Mehta Mar 21 '22 at 19:33
  • @WillJagy Can't say I understand your analogy, but I guess you might be hinting that there is an approach to finding the value of $f(10^{10})$ that may not generalize to all $n$. – Rushabh Mehta Mar 21 '22 at 19:48

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If $p(x)$ has your desired property, then $\frac{p(x)}{10^{10}}$ is a polynomial with leading coefficient $\frac{1}{10^{10}}$ that sends integers to integers. It's a theorem that all polynomials that send integers to integers are $\mathbb Z$-linear combinations of the polynomials $$ p_k(x) = \binom{x}{k} = \frac{x(x-1) \cdots (x-k+1)}{k!}. $$ In particular, in order to get any coefficient with a denominator of $10^{10}$ (or even a denominator divisible by $5^{10}$), we must use polynomials with degree $k$ large enough that $k!$ is divisible by $5^{10}$. This requires $k \geq 45$, so your example (which happens to equal $45! p_{45}(x+44)$) is optimal. The same approach generalizes to prove that for all $n$, $f(n)$ is the smallest $k$ such that $n | k!$.

Ravi Fernando
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