$\blacksquare~$ Problem: Find all $P(x)$ $\in$ $\mathbb{R}[x]$ such that for some $c \in \mathbb{R}$, the following functional equation holds \begin{align*} (x + 1)P(x - 1) - (x - 1)P(x) = c \end{align*}
$\blacksquare~$ My Attempt:
We will check at first for $x = 0 \text{ and } -1$ and by some simple calculation we will obtain that \begin{align*} P(0) = P( - 1) = \frac{c}{2} \end{align*}
Let's consider a polynomial $Q(x)$ $\in$ $\mathbb{R}[x]$ such that
\begin{align*}
Q(x) = P(x) - \frac{c}{2}
\end{align*}
Therefore our new functional equation becomes
\begin{align}
&(x + 1)Q(x - 1) - (x - 1) Q(x) = 0\\
\implies & (x + 1)Q(x - 1) = (x - 1) Q(x) \quad \quad \cdots\cdots (1)
\end{align}
Then we will make a claim about our new defined polynomial $Q(x)$.
Claim: The polynomial $Q(x)$ follows the following conditions \begin{align*} Q ( x ) = \begin{cases} 0 & \text{ if } Q \text{ is a constant polynomial in } \mathbb{R}[x] \\ b \cdot x ( x + 1 ) & \text{ otherwise and for some }b \in \mathbb{R} \\ \end{cases} \end{align*}
$\bullet~$ Proof: From the way we constructed $Q(x)$ we have \begin{align*} Q(0) = Q(-1) = 0 \end{align*} therefore let's consider the two cases separately
$\bullet~$ Case $1$: We have $Q$ as a constant polynomial. Say for some $a$ $\in$ $\mathbb{R}$ \begin{align*} Q(x) \equiv a \quad \text{for all } x \in \mathbb{R} \end{align*} Then, we have for any $x$ in $\mathbb{R}$ \begin{align*} &a \cdot (x + 1) = a \cdot (x - 1)\\ \implies & ax + a = ax - a\\ \implies & a = 0 \end{align*} Hence the first part of our claim is proved.
$\bullet~$ Case $2$: If $Q(x)$ is non-constant, then it's obvious that the polynomial $Q(x)$ has roots $0, -1,$ therefore, \begin{align*} Q(x) = x(x + 1) M(x) \quad \text{for some polynomial } M(x) \in \mathbb{R}[x] \end{align*} such that $\text{deg}(Q) > \text{deg}(M)$.
Therefore, we have from $(1)$ \begin{align*} \implies &(x - 1) \cdot x \cdot (x + 1) M(x - 1) = (x - 1) \cdot x \cdot (x + 1) M(x)\\ \implies & M(x - 1) = M(x) \quad \text{for all } x \in \mathbb{R} - \{-1, 0, 1\} \end{align*} Now on doing $x \mapsto x + 1$ we have \begin{align*} M(x) = M(x + 1) \quad \text{and similarly for each map } ( x + j ) \mapsto \{x + (j + 1)\} \end{align*} Hence we have the conclusion \begin{align*} &M(x) = M(x + 1) = M(x + 2) = \cdots = M(x + n) = \cdots\\ \implies & M(x) \equiv b \quad \text{for some }b \text{ in } \mathbb{R} \text{ and } \forall \text{ } x \in \mathbb{R} \end{align*}
again from the quotient form of $Q(x)$ we have \begin{align*} Q(x) = b \cdot x(x + 1) \quad \text{for some } b \text{ in } \mathbb{R} \end{align*} Hence we have proved the second part of our claim too.
Therefore we have proved both our claims!
Hence we obatin that the polynomial $Q(x)$ precisely. Which implies that our polynomial $P(x)$ will be \begin{align*} P ( x ) = \begin{cases} Q(x) + \frac{c}{2} = \frac{c}{2} & \text{ if } Q \text{ is a constant polynomial in } \mathbb{R}[x]\\ Q(x) + \frac{c}{2} = b \cdot x ( x + 1 ) + \frac{c}{2} & \text{ otherwise and for some }b \in \mathbb{R} \\ \end{cases} \end{align*}
Hence we are done!
Please check the solution, and please give some new ideas too :)