Canadian Mathematical Olympiad

The answer is $P(x)$ being any constant polynomial and $P(x) \equiv kx^2 + kx + c$ for any (nonzero) constant $k$ and constant $c$.
Let $\Lambda$ be the expression $(x+1)P(x-1) - (x-1)P(x)$, i.e. the expression in the problem statement.
Substituting $x = -1$ into $\Lambda$ yields $2P(-1)$ and substituting $x = 1$ into $\Lambda$ yields $2P(1)$. Since $(x+1)P(x-1) - (x-1)P(x)$ is a constant polynomial, $2P(-1) = 2P(0)$. Hence, $P(-1) = P(0)$.
Let $c = P(-1) = P(0)$ and $Q(x) = P(x) - c$. Then $Q(-1) = Q(0) = 0$. Hence, $0, -1$ are roots of $Q(x)$. Consequently, $Q(x) = x(x+1)R(x)$ for some polynomial $R$. Then $P(x) - c = x(x+1)R(x)$, or equivalently, $P(x) = x(x+1)R(x) + c$.
Substituting this into $\Lambda$ yield
$$(x+1)((x-1)xR(x-1) + c) - (x-1)(x(x+1)R(x) + c)$$
This is a constant polynomial and simplifies to
$$x(x-1)(x+1)(R(x-1) - R(x)) + 2c.$$

Since this expression is a constant, so is $x(x-1)(x+1)(R(x-1)-R(x))$. Therefore, $R(x-1)-R(x) = 0$ as a polynomial. Therefore, $R(x) = R(x-1)$ for all $x \in \mathbb{R}$. Then $R(x)$ is a polynomial that takes on certain values for infinitely many values of $x$. Let $k$ be such a value. Then $R(x) - k$ has infinitely many roots, which can occur if and only if $R(x) - k = 0$. Therefore, $R(x)$ is identical to a constant $k$. Hence, $Q(x) = kx(x+1)$ for some constant $k$. Therefore, $P(x) = kx(x+1)+c = kx^2+kx+c$.
Finally, we verify that all such $P(x) = kx(x+1)+c$ work. Substituting this into $\Lambda$ yields
$$(x+1)(kx(x-1)+c) - (x-1)(kx(x+1)+c) = kx(x+1)(x-1) + c(x+1) - kx(x+1)(x-1) - c(x-1) = 2c.$$
Hence, $P(x) = kx(x+1)+c = kx^2+kx+c$ is a solution to the given equation for any constant $k$. Note that this solution also holds for $k=0$. Hence, constant polynomials are also solutions to this equation. $\square$

As in Solution 1, any constant polynomial $P$ satisfies the given property. Hence, we will assume that $P$ is not a constant polynomial.
Let $n$ be the degree of $P$. Since $P$ is not constant, $n \ge 1$. Let
$$P(x) = \sum_{i=0}^{n} a_{i}x^{i},$$
with $a_n \neq 0$. Then
$$(x+1) \sum_{i=0}^{n} a_{i}(x-1)^{i} - (x-1) \sum_{i=0}^{n} a_{i}x^{i} = C,$$
for some constant $C$. We will compare the coefficient of $x^n$ of the left-hand side of this equation with the right-hand side. Since $C$ is a constant and $n \ge 1$, the coefficient of $x^n$ of the right-hand side is equal to zero. We now determine the coefficient of $x^n$ of the left-hand side of this expression.
The left-hand side of the equation simplifies to
$$x \sum_{i=0}^{n} a_{i}(x-1)^{i} + \sum_{i=0}^{n} a_{i}(x-1)^{i} - x \sum_{i=0}^{n} a_{i}x^{i} + \sum_{i=0}^{n} a_{i}x^{i}.$$

We will determine the coefficient $x^n$ of each of these four terms.
By the Binomial Theorem, the coefficient of $x^n$ of the first term is equal to that of $x (a_{n-1}(x-1)^{n-1} + a_n(x-1)^n) = a_{n-1} - \binom{n}{n-1}a_n = a_{n-1} - n a_n$.
The coefficient of $x^n$ of the second term is equal to that of $a_n(x-1)^n$, which is $a_n$.
The coefficient of $x^n$ of the third term is equal to $a_{n-1}$ and that of the fourth term is equal to $a_n$.
Summing these four coefficients yield $a_{n-1} - n a_n + a_n - a_{n-1} + a_n = (2-n)a_n$.
This expression is equal to 0. Since $a_n \neq 0$, $n = 2$. Hence, $P$ is a quadratic polynomial.
Let $P(x) = a x^2 + b x + c$, where $a, b, c$ are real numbers with $a \neq 0$. Then
$$(x+1)(a(x-1)^2 + b(x-1) + c) - (x-1)(a x^2 + b x + c) = C.$$
Simplifying the left-hand side yields
$$(b-a)x + 2c = 2C.$$
Therefore, $b - a = 0$ and $2c = 2C$. Hence, $P(x) = a x^2 + a x + c$. As in Solution 1, this is a valid solution for all $a \in \mathbb{R} \setminus \{0\}$. $\square$