Putnam 2010, Problem B-4

Find all pairs of polynomials \( p(x) \) and \( q(x) \) with real coefficients for which
\[
p(x)q(x+1) - p(x+1)q(x) = 1.
\]

Solution:
**First solution.** The pairs $(p,q)$ satisfying the given equation are those of the form $p(x) = ax + b$, $q(x) = cx + d$ for $a, b, c, d \in \mathbb{R}$ such that $bc - ad = 1$. We will see later that these indeed give solutions.

Suppose $p$ and $q$ satisfy the given equation; note that neither $p$ nor $q$ can be identically zero. By subtracting the equations
$$p(x)q(x+1) - p(x+1)q(x) = 1$$
$$p(x-1)q(x) - p(x)q(x-1) = 1,$$
we obtain the equation
$$p(x)(q(x+1) + q(x-1)) = q(x)(p(x+1) + p(x-1)).$$

The original equation implies that $p(x)$ and $q(x)$ have no common nonconstant factor, so $p(x)$ divides $p(x+1) + p(x-1)$. Since each of $p(x+1)$ and $p(x-1)$ has the same degree and leading coefficient as $p$, we must have
$$p(x+1) + p(x-1) = 2p(x).$$

If we define the polynomials $r(x) = p(x+1) - p(x)$, $s(x) = q(x+1) - q(x)$, we have $r(x+1) = r(x)$, and similarly $s(x+1) = s(x)$. Put
$$a = r(0), b = p(0), c = s(0), d = q(0).$$

Then $r(x) = a$, $s(x) = c$ for all $x \in \mathbb{Z}$, and hence identically; consequently, $p(x) = ax + b$, $q(x) = cx + d$ for all $x \in \mathbb{Z}$, and hence identically. For $p$ and $q$ of this form,
$$p(x)q(x+1) - p(x+1)q(x) = bc - ad,$$
so we get a solution if and only if $bc - ad = 1$, as claimed.

**Second solution.** (Communicated by Catalin Zara.) Again, note that $p$ and $q$ must be nonzero. Write
$$p(x) = p_0 + p_1 x + \cdots + p_m x^m$$
$$q(x) = q_0 + q_1 x + \cdots + q_n x^n$$
with $p_m, q_n \neq 0$, so that $m = \deg(p)$, $n = \deg(q)$. It is enough to derive a contradiction assuming that $\max\{m,n\} > 1$, the remaining cases being treated as in the first solution.

Put $R(x) = p(x)q(x+1) - p(x+1)q(x)$. Since $m + n \geq 2$ by assumption, the coefficient of $x^{m+n-1}$ in $R(x)$ must vanish. By easy algebra, this coefficient equals $(m-n)p_m q_n$, so we must have $m = n > 1$.

For $k = 1, \ldots, 2m-2$, the coefficient of $x^k$ in $R(x)$ is
$$\sum_{i+j > k, j > i} \left(\binom{j}{k-i} - \binom{i}{k-j}\right)(p_i q_j - p_j q_i)$$
and must vanish. For $k = 2m-2$, the only summand is for $(i,j) = (m-1,m)$, so $p_{m-1} q_m = p_m q_{m-1}$.

Suppose now that $h \geq 1$ and that $p_i q_j = p_j q_i$ is known to vanish whenever $j > i \geq h$. (By the previous paragraph, we initially have this for $h = m-1$.) Take $k = m + h - 2$ and note that the conditions $i + j > k$, $j \leq m$ force $i \geq h-1$. Using the hypothesis, we see that the only possible nonzero contribution to the coefficient of $x^k$ in $R(x)$ is from $(i,j) = (h-1,m)$. Hence $p_{h-1} q_m = p_m q_{h-1}$; since $p_m, q_m \neq 0$, this implies $p_{h-1} q_j = p_j q_{h-1}$ whenever $j > h-1$.

By descending induction, we deduce that $p_i q_j = p_j q_i$ whenever $j > i \geq 0$. Consequently, $p(x)$ and $q(x)$ are scalar multiples of each other, forcing $R(x) = 0$, a contradiction.

**Third solution.** (Communicated by David Feldman.) As in the second solution, we note that there are no solutions where $m = \deg(p)$, $n = \deg(q)$ are distinct and $m + n \geq 2$. Suppose $p,q$ form a solution with $m = n \geq 2$. The desired identity asserts that the matrix
$$\begin{pmatrix} p(x) & p(x+1) \\ q(x) & q(x+1) \end{pmatrix}$$
has determinant 1. This condition is preserved by replacing $q(x)$ with $q(x) - tp(x)$ for any real number $t$. In particular, we can choose $t$ so that $\deg(q(x) - tp(x)) < m$; we then obtain a contradiction.