2003 Vietnamese Mathematical Olympiad

We have: $$(x^3+3x^2+3x+2)P(x-1)=(x^3-3x^2+3x-2)P(x)\forall x\in \mathbb{R}(1)$$ $$\iff (x+2)(x^2+x+1)P(x-1)=(x-2)(x^2-x+1)P(x)\forall x\in \mathbb{R}(2)$$ By substituting $x=-2$ into (2), we get $0=-28\cdot P(-2)$, so $P(-2)=0$. By substituting $x=2$ into (2), we get $28\cdot P(1)=0$, so $P(1)=0$. Therefore, - By substituting $x=-1$ into (1), we get $0=-9\cdot P(-1)$, so $P(-1)=0$. - By substituting $x=1$ into (1), we get $9\cdot P(0)=0$, so $P(0)=0$. From these results, it follows that $$P(x)=(x-1)x(x+1)(x+2)Q(x)\forall x\in \mathbb{R},(3)$$ where $Q(x)$ is a polynomial in $x$ with real coefficients. $$\text{It implies that }P(x-1)=(x-2)(x-1)x(x+1)Q(x-1)\forall x\in \mathbb{R}.(4)$$ From (2), (3) and (4), we get: $$\begin{array}{rl}& (x-2)(x-1)x(x+1)(x+2)(x^2+x+1)Q(x-1)\\ & =(x-2)(x-1)x(x+1)(x+2)(x^2-x+1)Q(x)\forall x\in \mathbb{R}.\end{array}$$ $$\text{It follows that }(x^2+x+1)Q(x-1)=(x^2-x+1)Q(x)\forall x\ne 0,\pm 1,\pm 2.$$ As each side of this equality is a polynomial in $x$, we get $$(x^2+x+1)Q(x-1)=(x^2-x+1)Q(x)\forall x\in \mathbb{R}.(5)$$ Since $(x^2+x+1,x^2-x+1)=1$, it implies that $$Q(x)=(x^2+x+1)R(x)\forall x\in \mathbb{R},(6)$$ where $R(x)$ is a polynomial in $x$ with real coefficients, and so $$Q(x-1)=(x^2-x+1)R(x-1),\forall x\in \mathbb{R}.(7)$$ From (5), (6) and (7), we get: $$(x^2+x+1)(x^2-x+1)R(x-1)=(x^2-x+1)(x^2+x+1)R(x)\forall x\in \mathbb{R}.$$ But $(x^2-x+1)(x^2+x+1)\ne 0\forall x\in \mathbb{R}$, we have $R(x-1)=R(x)\forall x\in \mathbb{R}$, so $R(x)$ is a constant, and from (6), (3) we get: $$P(x)=c(x-1)x(x+1)(x+2)(x^2+x+1)\forall x\in \mathbb{R},$$ where $c$ is an arbitrary real constant. A direct verification and shows that these polynomials satisfy the given relation and thus they are polynomials sought for.