Cursed polynomials

Let the polynomial have two variables, $x$ and $y$ , such that it is of the form $p(x, y)$ . We will now compare degrees of both sides. Let $\deg_x (p)=d$ and $\deg_y (p)=e$ . We have $p(u, p(v,w)) = p(u+v, w)$ , so

$\begin{array}{c c c} & \text{LHS} & \text{RHS}\\ \deg(u) = & d & d\\ \deg(v) = & de & d\\ \deg(w) = & e^2 & e\\ \end{array}$

Equating degrees, we have $d = de$ , so $d=0$ or $e=1$ . We also have $e=e^2$ , so $e=0$ or $e=1$ .

If $e=0$ , we must have $d=0$ , so $p(x, y)=c$ . Combining this with $p(2016, 1)=4033$ , we get $p(x, y) = 4033$ , which satisfies the polynomial equation.

If $e=1$ , $p(x, y) = a(x) y + b(x)$ for $a, b \in \mathbb{R}[x]$ . Substituting this into the polynomial equation, we have

$\begin{aligned} p(u, p(v, w)) &= p(u+v, w)\\ p(u, a(v) w + b(v)) &= a(u+v) w + b(u+v)\\ a(u) (a(v) w + b(v)) + b(u) &= a(u+v) w + b(u+v)\\ a(u) a(v) w + a(u) b(v) + b(u) &= a(u+v) w + b(u+v)\\ \end{aligned}$

Equating similar parts results in

$\begin{aligned} a(u) a(v) &= a(u+v)\\ a(u) b(v) + b(u) &= b(u+v)\\ \end{aligned}$

Substituting $u=v$ in the first equation gives us $a(u)^2 = a(2u)$ , so either $a$ is constant or linear. If $a$ is constant, $b$ must be constant. If $a$ is linear, $b$ will follow to be either linear or constant, so $b(x)=cx$ .

Thus, substituting these results back into the original equation gives us either $p(x, y)=cx + y$ or $p(x, y)=c$ . Since $p(2016, 1) = 4033$ , we must have $p(x, y)=2x+y$ or $p(x, y) = 4033$ as the only possible polynomials which satisfy.

Let $u=v=0$ . Then $p\left(0,p(0,w)\right)=p(0,w)$ . If we then set $p(0,w)=r(w)$ , we get the identity $r\left(r(x)\right)=r(x)$ . If the degree of $r$ is at least $1$ , it must be a linear function when we compare the degrees of both sides. Setting $r(x)=ax+b$ and solving $r\left(r(x)\right)=r(x)$ gives $r(x)=x$ or that $r(x)$ is constant.

Case 1: $r(x)=x$

This implies that $p(x,y)=y+xq(x,y)$ for some real multivariate polynomial $q$ . Substituting this result to the given condition yields $vq(v,w)+uq\left(u,w+vq(v,w)\right)=(u+v)q(u+v,w).$ Let $\text{deg}_xq(x,y)=d_x$ and $\text{deg}_yq(x,y)=d_y$ . Comparing the degree of $v$ in the last identity above we get $d_y(d_x+1)=0$ . This tells us that $q(x,y)$ is independent of $y$ . Setting $q(x,y)=Q(x)$ simplifies the identity to $Qq(v)+uQ(u)+(u+v)Q(u+v)$ .Let $u=v$ and we obtain $Q(u)=Q(2u)$ for any nonzero $u$ , hence $Q$ must be constant. $p(2016,1)=4033\Rightarrow Q(x)=2\Rightarrow p(x,y)=x+2y$ .

Case 2: $r(x)=c$ where $c$ is some constant.

Let $u=0$ . Then $p(v,w)=p(0,p(v,w))=c$ . This $c$ has to be $4033$ , so $p(x,y)=4033$ is the only other solution.

Hence there are $2$ possible polynomials $p$ .