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?
Best Answer
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 $P(x) = x$.