Find all polynomials such that $P (x) = (x-P (0))(x-P (1))...(x-P (n-1))$ Find all polynomials $P(x)$ of degree $n>0$ with integer coefficients such that, for every real number $x$, we have:
$$P(x) = (x-P (0))(x-P (1))...(x-P (n-1)) $$
I thought that i proved that $P(0) = 0$ but actually the conclusion was wrong, as showed in the comments. I'm back to the incial point.
I don't now how to proceed, can somebody help me?
 A: 1. If $n=1$, it's easy to see that the only solution is $P(x)=x$.
2. If $n \geq 2$, set $x=0$:
$$P(0)=(-1)^n\cdot P(0)\cdot P(1)\cdot \ldots \cdot P(n-1)$$
If $P(0)$ is non-zero, we have
$$ P(1)\cdot P(2) \cdot \ldots \cdot P(n-1)=(-1)^{n+1}$$
This means that all the values $P(1), P(2), \ldots, P(n)$ are $\pm 1$. However, none of this values can be $1$, because by setting $x = 1$, we would have $P(1)=0$. This means they are all $-1$, so 
$$P(x)=[x-P(0)]\cdot (x+1)^{n-1}$$ 
Now setting $x=1$ we get 
$$-1=P(1)=2^{n-1}\cdot [1-P(0)]$$
which is false. This means we must have $P(0) = 0$. Therefore the given relation writes as
$$P(x)=x(x-P(1))(x-P(2))\cdot\ldots \cdot(x-P(n-1))$$
Now setting $x=1$ gives 
$$(1-P(1))(1-P(2))\cdot \ldots \cdot(1-P(n-1))=P(1) $$
However, since $\gcd(P(1),1-P(1))=1$, we must have either $P(1)=0$ or $P(1)=2$. 
If $P(1) = 0$, we have $P(i)=1$ for some $2\leq i \leq n-1$, but setting $x=i$ gives
$$1=i^2(i-1)(i-P(2))\cdot \ldots \cdot(i-P(n-1))$$ 
which is false again.
If $P(1) = 2$ we get 
$$(1-P(2))\cdot \ldots \cdot(1-P(n-1))=-2$$
None of these factors can be $1$ (similar reasoning as previously), so $P(i)=3$ for some $i$ and $P(j)=0$, for $2\leq j \leq n,\ j\neq i$. We get $P(x)=x^{n-2}(x-2)(x-3)$, which is not a solution.
In conclusion, the only solution is $P(x) = x$.
A: First of all, note that if $P(0)\ne 0$, one has $$\prod_{k=1}^{n-1}P(k)=(-1)^n,$$ so that even number of $P(k)$'s $(k\ge 1)$are $1$ and the rest are $-1$, so that $P(x)$ can be of the form $(x-P(0))(x-1)^e(x+1)^{n-1-e}$, for some $e=2,4,6,\cdots$. However, in that case, $P(k)=\pm 1$ for $k=1,2,\cdots,n-1$, and therefore putting $x=k, k\in \{1,2,\cdots,n-1\}$ gives $$\pm 1=(k-P(0))(k-1)^e(k+1)^{n-1-e},$$ which is absurd. Therefore, one must have $P(0)=0$. Consequently, $$P(x)=x(x-P(1))\cdots(x-P(n-1)).$$ Now note that for any $k\ge 1$, $$P(k) = k(k-P(1))\cdots(k-P(k))\cdots(k-P(n-1))\\\implies k|P(k),\ k-P(k)|P(k).$$ Therefore, there are integers $r,s$ such that $P(k)=kr,\ P(k)=s(k-P(s))$. Therefore, $kr=s(k-kr)\implies s(1-r)=r\implies 1-r=\pm 1\implies r=0,2$ and $s=0,-2$. Hence, either $P(k)=0$, or $P(k)=2k$. Now, let $P(k)=2k$ for some $k\ge 1$. Then, $$2k=P(k)=k(k-P(1))\cdots(k-P(k))\cdots(k-P(n-1))\\\implies -2=k(k-P(1))\cdots(k-P(k-1))\cdots (k-P(n-1))\\\implies k=1,2.$$ However, in that case, $P(k)=0$ for all $k\ge 3$, so that $$P(x)=x^{n-2}(x-P(1))(x-P(2))\implies 0=P(3)=3^{n-2}(3-P(1))(3-P(2)),$$ which yields a contradiction. Therefore, one must have, $P(k)=0$ for all $k\ge 1$. Therefore, $P(x)=x^n$. However, this works only when $n=1$. Therefore, the only solution is $P(x)=x$.
