Putnam 2005, Problem B-2

Find all positive integers $n, k_1, \ldots, k_n$ such that $k_1 + \cdots + k_n = 5n - 4$ and $\frac{1}{k_1} + \cdots + \frac{1}{k_n} = 1$.

Solution:
By the arithmetic-harmonic mean inequality or the Cauchy-Schwarz inequality,
\[
(k_1 + \cdots + k_n) \left( \frac{1}{k_1} + \cdots + \frac{1}{k_n} \right) \geq n^2.
\]
We must thus have \(5n - 4 \geq n^2\), so \(n \leq 4\). Without loss of generality, we may suppose that \(k_1 \leq \cdots \leq k_n\).

If \(n = 1\), we must have \(k_1 = 1\), which works. Note that hereafter we cannot have \(k_1 = 1\).

If \(n = 2\), we have \((k_1,k_2) \in \{(2,4),(3,3)\}\), neither of which work.

If \(n = 3\), we have \(k_1 + k_2 + k_3 = 11\), so \(2 \leq k_1 \leq 3\). Hence
\[
(k_1,k_2,k_3) \in \{(2,2,7),(2,3,6),(2,4,5),(3,3,5),(3,4,4)\},
\]
and only \((2,3,6)\) works.

If \(n = 4\), we must have equality in the AM-HM inequality, which only happens when \(k_1 = k_2 = k_3 = k_4 = 4\).

Hence the solutions are \(n = 1\) and \(k_1 = 1\), \(n = 3\) and \((k_1,k_2,k_3)\) is a permutation of \((2,3,6)\), and \(n = 4\) and \((k_1,k_2,k_3,k_4) = (4,4,4,4)\).

Remark: In the cases \(n = 2,3\), Greg Kuperberg suggests the alternate approach of enumerating the solutions of \(1/k_1 + \cdots + 1/k_n = 1\) with \(k_1 \leq \cdots \leq k_n\). This is easily done by proceeding in lexicographic order: one obtains \((2,2)\) for \(n = 2\), and \((2,3,6),(2,4,4),(3,3,3)\) for \(n = 3\), and only \((2,3,6)\) contributes to the final answer.