Mathematical Reflections 2017, Issue 2 - Problem U403

Problem:
Find all cubic polynomials $P(x) \in \mathbb{R}[x]$ such that $$P\left(1-\dfrac{x(3x+1)}{2}\right)-P^2(x)+P\left(\dfrac{x(3x-1)}{2}-1\right)=1 \qquad \forall x \in \mathbb{R}.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let $P(x)=ax^3+bx^2+cx+d$ be a third-degree polynomial with $a,b,c,d \in \mathbb{R}$, $a \neq 0$, such that
$$P\left(1-\dfrac{x(3x+1)}{2}\right)-P^2(x)+P\left(\dfrac{x(3x-1)}{2}-1\right)=1 \qquad \forall x \in \mathbb{R}.$$
Setting $x=-1,x=0,x=1$ into the given relation, we obtain
$$\begin{array}{rcl} P(0)-P^2(-1)+P(1)&=&1 \\ P(1)-P^2(0)+P(-1)&=&1 \\ P(-1)-P^2(1)+P(0)&=&1. \end{array}$$
Adding these three equations side by side, we get $$(P(-1)-1)^2+(P(0)-1)^2+(P(1)-1)^2=0,$$ so $P(-1)=P(0)=P(1)=1$. It follows that
$$\begin{array}{rcl} d&=&1 \\ a+b+c+d&=&1 \\ -a+b-c+d&=&1. \end{array}$$ We obtain $d=1,b=0$ and $c=-a$. So, $P(x)=ax^3-ax+1$.
Setting $x=2$ into the given relation, we have $$P(-6)-P^2(2)+P(4)=1,$$ which gives $$-36a^2-162a+1=1.$$ Likewise, setting $x=-2$ into the given relation, we have $$P(-4)-P^2(-2)+P(6)=1,$$ which gives $$-36a^2+162a+1=1.$$ Subtracting, we get $264a=0$, i.e. $a=0$. So, we obtain $P(x)=1$.