Polynomial and number theory problem $n\mid P(P(n)) - (n-1)$

Question

Let $P(x)$ be an integer polynomial such that for all $n \in \mathbb{Z^+}$: $$n\mid P\left(P(n)\right) - (n-1).$$ Prove that there is no $x_0 \in \mathbb{R}$ such that $P\left(x_0\right)=0.$

Here is what I got:

Because $n\mid P\left(P(n)\right) - (n-1)\Rightarrow n\mid P(\left(P(n)\right)+1,\forall n \in \mathbb{Z^+}.$

On the other hand, we have: $n\mid P(P(n)) - P(P(0)),\forall n \in \mathbb{Z^+}.$

Thus $n\mid P(P(0)) + 1,\forall n \in \mathbb{Z^+}.$

This leads to $P(P(0)) = -1.$

Moreover, we have $2\mid P(P(2)) + 1\Rightarrow P(P(2))$ is odd.

However, we know that $P(x) = a_nx^n + a_{n-1}x^{n-1} +\dots + a_1x+ 1$ and $P(2) = 2A + 1 \Rightarrow P(P(2)) = n(2A+1)B+1$ which is odd, so $n$ should be odd.

Answer 1

If what needs to be proven is: there is no $x_0 \in \color{red}{\mathbb{R}}$ such that $P(x_0)=0$:

Counter example:

Let $P(x) = x^2 + x-1$ which has two real roots $x = (-1\pm \sqrt{5})/2$.

However, $$P(P(n))-(n-1) = n (-2 + 2 n^2 + n^3) \equiv0 \pmod n \hspace{1cm} \forall n \in \mathbb{Z}^+.$$

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If what needs to be proven is: there is no $x_0 \in \color{red}{\mathbb{Z}}$ such that $P(x_0)=0$:

Denote $P(x)=\sum_{k=1}^Ka_k x^k+ a_0$. It's easy to prove that $$\begin{align} &P(P(n)) = \sum_{k=1}^Ka_k \left(n\sum_{k=1}^Ka_k n^{k-1}+ a_0 \right)^k+ a_0 \equiv \sum_{k=1}^Ka_k a_0^k+ a_0 \pmod n\hspace{0.5cm} \\ \implies& \sum_{k=1}^Ka_k a_0^k+ a_0 +1 \equiv0\pmod n\hspace{1cm} \forall n\\ \implies& a_0\left(\sum_{k=1}^Ka_k a_0^{k-1}+ 1\right) +1 \equiv0\pmod n \hspace{1cm} \forall n\\ \implies& a_0\left(\sum_{k=1}^Ka_k a_0^{k-1}+ 1\right) +1 =0 \\ \implies& a_0\left(P(a_0)-a_0+ 1\right) +1 =0. \tag{1} \end{align}$$

From $(1)$, we deduce that $a_0 \in \{1,-1\} $ and $$P(a_0)-a_0 \equiv 0 \pmod 2. \tag{2}$$

If $a_0 = 1$:

From $P(x_0) = 0 $, we deduce that $ x_0\left(\sum_{k=1}^Ka_k x_0^{k-1}\right) + 1 = 0 \implies x_0 \in \{1,-1 \}$.

As $(2) \iff P(1) \equiv 1 \pmod 2$, $x_0$ must be equal to $-1$, that is: $P(-1) =0. $ We deduce that $$1+0 \equiv P(1) + P(-1) = 2\sum_{k \text{ even}}a_k \equiv 0 \pmod 2 \implies \text{contradiction}.$$

If $a_0 = -1$:

From $P(x_0) = 0 $, we deduce that $ x_0\left(\sum_{k=1}^Ka_k x_0^{k-1}\right) - 1 = 0 \implies x_0 \in \{1,-1 \}.$

As $(2) \iff P(-1) \equiv -1 \pmod 2$, $x_0$ must be equal to $1$, that is: $P(1) =0.$ We deduce that $$-1+0 \equiv P(-1) + P(1) = 2\sum_{k \text{ even}}a_k \equiv 0 \pmod 2 \implies \text{contradiction}.$$

We can conclude that there is no $x_0 \in \color{red}{\mathbb{Z}}$ such that $P(x_0)=0$.