$P(\lfloor x\rfloor)=\lfloor P(x)\rfloor$

Question

Find all polynomials with real coefficients $P(X)$ such that $P(\lfloor x\rfloor)=\lfloor P(x)\rfloor$ for all real x.

I've asked a similar question with the fractional part function before instead, and I thought of changing the fractional part to a floor function. The thing is, other than substituting values like $0, 1$ etc. I genuinely can't think of any strategies to work on this one. Any hints or solutions would be highly appreciated.

Answer 1

Hints: Fill in the gap yourself.

1. The crux of the proof is the idea that for any integer $n$ and fractional part $ 0 \leq \alpha < 1$, we must have $ 0 \leq f(n + \alpha ) - f(n) < 1. $
2. Show that if $P(x)$ is a polynomial of degree $ \geq 2$, then there exists a large enough $n$ such that $ |P(n + 0.5 ) - P(n)| > 2$.
3. Hence, a polynomial that satisfies the conditions has degree at most 1.
4. Show that if $P(x) = ax + b$, then we must have $ a, b \in \mathbb{Z}$.
5. Show that $ a \in \{0, 1 \}$.
6. Hence, conclude that $ P (x) $ must be of the form $ ax + b$ where $ a \in \{0, 1 \}, b \in \mathbb{Z}$.

Answer 2

Note that such a $P$ must map integers to integers, since $P(n) = P(\lfloor n\rfloor) = \lfloor P(n)\rfloor$. If $P(n)=k$, the given property forces $P(x) \in [k,k+1)$ for all $x\in[n,n+1)$; by continuity, we must then have $P(n+1)=k$ or $P(n+1)=k+1$. In particular, the difference polynomial $P(x+1)-P(x)$ either equals $0$ infinitely often (hence is the constant $0$) or equals $1$ infinitely often hence is the constant $1$). It follows quickly that $P(x)=m$ and $P(x)=x+m$, for integers $m$, are the only possible solutions (and they are indeed solutions).

Answer 3

1. For every integer $n$, we have $P(n) \in \Bbb Z$.

   This is clear by putting $x = n$ in the formula.

2. For every integer $n$, we have $P(n + 1) - P(n) = 0$ or $1$.

   This is because for all $\epsilon \in [0, 1)$, the value of $P(n + \epsilon)$ lies in the interval $[P(n), P(n) + 1)$, and $P$ is continuous.

3. $P$ has degree $\leq 1$.

   Otherwise $|P(n + 1) - P(n)|$ would be too big for large enough $n$.

Thus $P(x) = ax + b$, and by 2. we see that $a = 0$ or $1$.

So the only possible solutions are:

• $P(x) = x + b$ for some integer $b$, or
• $P(x) = b$ for some integer $b$.