Exercise 8, § 6

Algebra, IV, p. 100

Put

s_k = \sum A_{m_1 \ldots m_n} p_1^{m_1} \cdots p_n^{m_n},

where the $A_{m_1 \ldots m_n}$ are rational numbers and the sum is extended over all families $(m_1, \ldots, m_n)$ such that $\sum i \cdot m_i = k$ . Show that $\sum A_{m_1 \ldots m_n} = 0$ for $k > 1$ and $\sum |A_{m_1 \ldots m_n}| = 1$ for $k \geq 1$ . Deduce that on writing $q_j = -p_j$ we have

(-1)^k s_k = \sum |A_{m_1 \ldots m_n}| q_1^{m_1} \cdots q_n^{m_n}.

Solution

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