Find all polynomials $P(x)$ with integer coefficients such that for all real numbers $s$ and $t$, if $P(s)$ and $P(t)$ are both integers, then $P(s+t)$ is also an integer.

This is a problem inspired by problem $5$ of APMO $2018$.

Without loss of generality, we may assume $$P(x) = \displaystyle \sum_{i=0}^n a_i x^i, \, a_n>0.$$

This is because if $P(x)$ is a solution, then so is $-P(x)$.

However, I can't even guess what kind of polynomials could be other than integer constant.

  • 1
    $\begingroup$ Also true for all degree 1 polynomials. E: Also your example is not valid: polynomials in the question have integral coefficients. $\endgroup$ – Mees de Vries Aug 10 '18 at 7:56
  • $\begingroup$ Thank you, I've edited the post. $\endgroup$ – Star Chou Aug 10 '18 at 8:07
  • 1
    $\begingroup$ hint: if $n>1$, and if $K$ is sufficiently big, then there are $s$'s near $K$, with spaces between these $s$'s approximately $1/(na_nK^{n-1})$ ($na_nK^{n-1}$ is approximately $P'(K)$), such that $P(s)$ is an integer. The same works around $-K$ - let's call these solutions $t$'s. But then $s+t$'s are around $0$, and there are arbitrary many of them (when you take bigger and bigger $K$), which is impossible. So $n\leq 1$. $\endgroup$ – user8268 Aug 10 '18 at 8:14

This should work, although it's definitely not as clean as the solution suggested by user8628 in the comments on the original question. It starts off similar to one solution to the APMO problem and then gets weird. But here it is anyway:

Say we have an example where $d=\deg(P)\geq 2$. For any real $s$ for which $P(s)\in\mathbb{Z}$, we necessarily have $P(ns)\in\mathbb{Z}$ for all positive integers $n$. Let $d$ be the degree of the polynomial, and set

$$x_n = P(ns) = \sum_{i=0}^d \left(a_i s^i\right) n^i = \vec{v}\cdot\vec{p_n},$$


$$\vec{v} = \left[a_0s^0,\cdots,a_ds^d\right],\ \ \vec{p_n} = \left[n^0, n^1,\cdots, n^d\right].$$

So, taking the Vandermonde matrix $M$ determined by $\alpha_n = n$ of degree $d+1$, we can write

$$\left[x_0,\cdots,x_d\right] = M\vec{v}.$$

As this matrix is invertible and has integer entries, its inverse $M^{-1}$ has rational entries, so we can write our vector $\vec{v}$ as a vector with rational entries with denominators dividing $N=\det(M)$, which implies that $s^i$ is rational for all $i$; in particular, for $i=1$ (since all constant polynomials work), we have that $s^d$ is rational, and in particular an integer multiple of $1/N'$, with $N'=1/(Na_d)$.

Now, let


We have that $Q$ is of degree $\leq d-1$ and all the reals $s$ for which $Q(s)\in\mathbb{Z}$ also satisfy $P(s)\in\mathbb{Z}$. Now, we see that, if $\deg(Q) > 0$, there are asymptotically $cx^{\deg(Q)}$ such numbers in the interval $[0,x]$, while there are asymptotically $a_dx^d$ in the interval $[0,x]$. This is a contradiction, so $\deg(Q) = 0$ and thus


for some integers $a,b,c$ with $b>1$. Now, let

$$s = \left(\frac{1}{a}\right)^{1/b},\ \ t = \left(\frac{2}{a}\right)^{1/b},$$

so $P(s)=1+c,P(t)=2+c$. We see

$$s+t = \frac{1+2^{1/b}}{a^{1/b}},$$


$$P(s+t) = \left(1+2^{1/b}\right)^b+c.$$

However, if this is some integer $m+c$, we have


which cannot hold for $b\geq 2$, as it would imply that, looking at minimal polynomials,

$$(x-1)^b-2 \big| x^b-m \implies (x-1)^b-2 = x^b - m,$$

which gives, looking at the coefficients of $b-1$,


a clear contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.