Find all polynomials p and q with real coefficients so that $p(x)q(x+1)-p(x+1)q(x)=1$ identically.
Let $R(x)=p(x)q(x+1)-p(x+1)q(x)$.
If one can show that both $p$ and $q$ are at most linear, the problem is relatively straightforward. So assume for a contradiction that one of $p$ and $q$ has degree at least 2. Note that if $m$ and $n$ are the degrees of p and q respectively, then the coefficient of $x^{m+n-1}$ of $p(x)q(x+1)-p(x+1)q(x)$ is $(m-n)p_mq_n,$ so in order for $R(x)$ to equal 1 identically, we must have that $m=n$.
From an online source, it seems that the coefficient of $x^k$ in $R(x) $ is equal to $\sum_{i+j > k, j > i} ({j\choose k-i} - {i\choose k-j}) (p_iq_j-p_jq_i),$ and I know that the coefficient is equal to $\sum_{i=0}^k p_i [x^{k-i}] q(x+1) - \sum_{i=0}^k q_i [x^{k-i}] p(x+1) = \sum_{i=0}^k p_i (\sum_{j\ge k-i} {j\choose k-i} q_j) - \sum_{i=0}^k q_i (\sum_{j\ge k-i} {j\choose k-i} p_j),$ but I'm not sure how to rearrange this formula.
The coefficient of $x^{2m-2}$ equals $p_{m-1} (q_{m-1} +mq_m)+ p_{m-2}q_m - (p_{m}q_{m-2} + (p_{m-1}+mp_m)q_{m-1}) = mp_{m-1}q_m - mp_mq_{m-1} + p_{m-2}q_m - p_m q_{m-2}$. Does $p_{m-2}q_m = p_m q_{m-2}$?
I'm looking for a solution that proves that $p_i q_j = p_jq_i$ whenever $j > i\ge 0$ (e.g. by backward induction). If this solution isn't applicable, then I'd accept any solution that isn't identical to the following solution:
$R(x)-R(x-1) = 0\Rightarrow p(x)(q(x+1)+q(x-1)) - q(x)(p(x+1)+p(x-1))= 0.$ Since $p$ and $q$ are coprime by Bezout's identity, $p(x)$ divides $p(x+1)+p(x-1)$ and $q(x)$ divides $q(x+1)+q(x-1)$. Matching leading coefficients gives $2p(x)=p(x+1)+p(x-1), 2q(x)=q(x+1)+q(x-1).$ So $r(x) := p(x+1)-p(x)$ is equal to zero for every $x$ in $\mathbb{Z}$ by induction and similarly $s(x) := q(x+1)-q(x)$ is equal to zero for every $x\in \mathbb{Z}$. So $p$ and $q$ are at most linear (for if for instance $p$ had degree at least 2, then the coefficient of $x^{m-1}$ in $p(x+1)-p(x)$ would be $m\neq 0$). Hence substituting $p(x) = r(0) x + p(0), q(x) = s(0) x + q(0)$ into the original equation, we see that $p(0)s(0)-r(0)q(0)=1$, which gives all solutions.