If $P(a)=0 \Rightarrow P(a+1)=1$ then $P(x)$ has no repeated roots. 
Let $P(x) \in \mathbb{R}[x]$ be polynomial with all real roots and has the property that $P(a)=0 \Rightarrow P(a+1)=1$ for all $a \in \mathbb{R}$. Prove that $P(x)$ has a repeated root.

I think this problem statement is not true because if we suppose that $P(x)=x$ then $P(0)=0 \Rightarrow P(1)=1$, $\;P(x)$ has no repeated root.
Please suggest.
 A: Let us suppose that $P \in \mathbb{C}[x]$ has degree $k$, i.e., it has complex roots $\alpha_1,\ldots,\alpha_k$ with
$$
P(x)=c\prod_i(x-\alpha_i)
$$
where the $\alpha_i$ are distinct complex and $c\neq 0$. In addition, we know that
$$
P(x)-1=c\prod_i(x-\alpha_i-1).
$$
In particular, the coefficient of $x^{k-1}$ of $P(x)$ verifies
$$
-c\sum_i \alpha_i = - c\sum_i (\alpha_i+1).
$$
This is impossible whenever $k\ge 2$.

Edit: The condition of $\alpha_i$ being distinct is necessary. Indeed, the polynomial $P(x):=x^k$ has all roots real and equal to $0$, and $P(a)=0 \implies P(a+1)=1$ for all $a \in \mathbf{R}$.
A: You need the degree of $P$ to be $d\ge2$. If so then
$$P(x)=(x-a_1)(x-a_2)\ldots(x-a_d)$$
for distinct $a_i$. Then
$$P(x)-1=(x-a_1-1)(x-a_2-1)\ldots(x-a_d-1).$$
Can you get a contradiction from these?
A: For $n>1$ we have $P(x)=(x-a)^n$ then $P(a+1)=(a+1-a)^n=1$
If $P(x)$ has no repeated root then it doesn't happen in general
Anyway  carat, the original poster, is right: for $n=1$ it is false. It even makes no sense talking of multiple root for $n=1$
Suppose the root are $3$ then we have
$P(x)=(x - a) (x - b) (x - c)$
Use the condition that $P(a+1)=1;\;P(b+1)=1;\;P(c+1)=1$
$$
\left\{ \begin{gathered}
  \left( {a + 1 - a} \right)\left( {a + 1 - b} \right)\left( {a + 1 - c} \right) = 1 \hfill \\
  \left( {b + 1 - a} \right)\left( {b + 1 - b} \right)\left( {b + 1 - c} \right) = 1 \hfill \\
  \left( {c + 1 - a} \right)\left( {c + 1 - b} \right)\left( {c + 1 - c} \right) = 1 \hfill \\ 
\end{gathered}  \right.
$$
which simplifies to
$$\left\{ \begin{gathered}
  \left( {a - b + 1} \right)\left( {a - c + 1} \right) = 1 \hfill \\
  \left( {b - a + 1} \right)\left( {b - c + 1} \right) = 1 \hfill \\
  \left( {c - a + 1} \right)\left( {c - b + 1} \right) = 1 \hfill \\ 
\end{gathered}  \right.$$
which is verified when $a=b=c$
Therefore $x=a$ is a triple root. $P(x)=(x-a)^3$
In a similar way it can be proved for any degree
