Putnam 2014, Problem A-5

Let
$$P_n(x) = 1 + 2x + 3x^2 + \cdots + nx^{n-1}.$$
Prove that the polynomials $P_j(x)$ and $P_k(x)$ are relatively prime for all positive integers $j$ and $k$ with $j \neq k$.

Solution:
First solution: Suppose \(P_i(z) = P_j(z) = 0\) for \(i \neq j\) and \(z \in \mathbb{C}\). Then \(z\) cannot be nonnegative real (else \(P_i(z), P_j(z) > 0\)). Let \(w = z^{-1} \neq 0,1\). For \(n \in \{i+1, j+1\}\), \(w^n = nw - n + 1\). Then \(w^{i+1} - (i+1)w + i = 0\) and \(w^{j+1} - (j+1)w + j = 0\). Subtracting, \((w-1)(w^i + w^{i-1} + \dots + 1 - (i+1)) = 0\), so \(w^i + \dots + 1 = i+1\). But \(|w^i + \dots + 1| \leq |w|^i + \dots + 1 < i+1\) if \(|w| < 1\), and \(> i+1\) if \(|w| > 1\), a contradiction.  
Third solution: (by David Feldman) Note \(P_n(x)(1 - x) = 1 + x + \dots + x^{n-1} - nx^n\). If \(|z| \geq 1\), then \(n|z|^n \geq |z|^{n-1} + \dots + 1 \geq |z^{n-1} + \dots + 1|\), with equality only if \(|z| = 1\) and \(z\) is positive real. Thus \(P_n(z)(1 - z) = 0\) has no solutions with \(|z| \geq 1\) except \(z = 1\). Since \(P_n(x)(1 - x)^2 = 1 - (n+1)x^n + nx^{n+1}\), it suffices to check \(C_n = \{z \in \mathbb{C} : 0 < |z| < 1, |z|^n |n + 1 - zn| = 1\}\) are pairwise disjoint. Let \(z = u + iv\) with \(v \geq 0\), and define \(E_z(n) = n \log |z| + \log |n + 1 - zn|\). Then \(E_z''(n) < 0\) iff \(\frac{u - v - 1}{(1 - u)^2 + v^2} < n < \frac{u + v - 1}{(1 - u)^2 + v^2}\). Also, \(E_z(0) = 0\) and \(E_z'(0) \geq 0\), so \(E_z(n) = 0\) has at most one solution with \(n > 0\).