Given a single variable real polynomial $p(x)$ of degree $n$ that maps integers to integers, show that $n!\cdot p(x)$ has integer coefficients.

I prefer an elementary solution if possible. It seems like some big machinery to take care of this should exist as well but I don't know any names.

For example, if $n=4$, some polynomial like $p(x)=\frac{(x-3)(x-8)(x-50)(x-13)}{24}$ works since $3,8,50,13$ are congruent to $3,0,2,1$ mod $4$. Any integer linear combinations of polynomials of this form works. But I can't say that these are the only possibilities.

What I typed earlier in the preceding paragraph is not correct. I meant polynomials such as $p(x)=\frac{(x-4)(x-5)(x-6)(x-7)}{24}$ works since the product of any $k$ consecutive integers is divisible by $k!$.

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    $\begingroup$ I just found the following which is related but gives no proof. math.stackexchange.com/questions/455041/… $\endgroup$
    – domoremath
    Sep 5, 2013 at 0:38
  • $\begingroup$ The claim made in the third sentence of that post would, if I'm not mistaken, imply this result. That might be the applicable "big machinery". $\endgroup$ Sep 5, 2013 at 1:16

2 Answers 2


Two words: Finite differences.

If $q$ is a polynomial, define another polynomial $\Delta q$ via: $(\Delta q)(x) = q(x+1)-q(x)$. Clearly, $\Delta^k p(0)$ is an integer for all $k$.

Then we know that $$p(x)=\sum_{k=0}^n \frac{\Delta^k p(0)}{k!}(x)_k$$


$$(x)_k = x(x-1)\cdots(x-k+1)$$

A similar proof, but with some proof of the finite difference result, is to note that the polynomial $(x)_k$ is of degree $k$, and we have $(x+1)_k-(x)_k=k(x)_{k-1}$. So every polynomial can be written uniquely as:

$$p(x)=\sum_{k=0}^{n} a_k(x)_k$$ where $n=\deg p(x)$. We then prove by induction that $a_k \cdot k!$ must be an integer.

If $p:\mathbb Z\to\mathbb Z$, we always have $a_0=p(0)\in\mathbb Z.$

Now $\Delta p:\mathbb Z\to\mathbb Z$, and:

$$\Delta p(x)=p(x+1)-p(x)=\sum_{k=0}^{n} ka_k(x)_{k-1}.$$

Since $p$ is smaller degree, by induction, you have that $k!a_k=(k-1)!ka_k\in\mathbb Z$ for each $k>0.$


I can prove the $n=2$ case with elementary math.

We're saying that $p$ maps integers to integers. That means that $f=2!\cdot p$ maps integers to multiples of $2$. Let's write $f(n)=\frac{r_1}{s_1}n^2 + \frac{r_2}{s_2}n + \frac{r_3}{s_3}$, and we can assume that each fraction has been written in lowest terms. We want to show that $s_1, s_2, s_3$ all equal $1$.

Since $f(0)$ is an integer, we immediately have that $s_3=1$.

Since $f(1)$ is an integer, we have that $\frac{r_1}{s_1} + \frac{r_2}{s_2}$ is an integer, so $s_1=s_2$. Let's just call it $s$.

Since $f(1)$ and $f(2)$ are both even integers, we have that $r_1 + r_2$ and $4r_1 + 2r_2$ are both multiples of $2s$. Subtracting twice the first expression from the second one, we obtain that $2s$ also divides $2r_1$, which implies that $s|r_1$. However, $r_1$ and $s$ are relatively prime, so $s=1$, q.e.d..

I don't immediately see how this type of argument would gracefully extend to higher degree polynomials, but it might.

  • $\begingroup$ I suppose if I want to be very rigorous, I would need to show that the $\frac{r_i}{s_i}$ coefficients can't be irrational. $\endgroup$
    – domoremath
    Sep 5, 2013 at 1:18
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    $\begingroup$ That's clear by solving a linear system. You can solve for the coefficients of a degree $n$ polynomial by knowing $n+1$ values of it. Cramer's rule show's that the solutions of this system, i.e. the coefficients of your polynomial, are rational. $\endgroup$ Sep 5, 2013 at 1:22

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