IMO 1979 · Problem 1

Let \(p\) and \(q\) be natural numbers such that \[ \frac pq=1-\frac12+\frac13-\frac14+\cdots-\frac1{1318}+\frac1{1319}. \] Prove that \(p\) is divisible by 1979.

Official text of the problem. Official problem paper (PDF)

Solution

Subtracting twice the sum of the even-indexed terms from the harmonic sum gives \[ \frac pq =\sum_{k=1}^{1319}\frac1k -2\sum_{j=1}^{659}\frac1{2j} =\sum_{k=1}^{1319}\frac1k-\sum_{k=1}^{659}\frac1k =\sum_{k=660}^{1319}\frac1k. \] Pair the denominators \(k\) and \(1979-k\). Since these pairs partition the integers from \(660\) to \(1319\), we obtain \[ \frac pq =\sum_{k=660}^{989}\left(\frac1k+\frac1{1979-k}\right) =1979\sum_{k=660}^{989}\frac1{k(1979-k)}. \]

Set \[ D=\prod_{k=660}^{1319}k. \] For every \(660\le k\le989\), the two distinct integers \(k\) and \(1979-k\) occur among the factors of \(D\). Consequently, \[ A=\sum_{k=660}^{989}\frac{D}{k(1979-k)} \] is an integer, and the preceding identity yields \[ pD=1979qA. \]

Now \(1979\) is prime. Indeed, \(\sqrt{1979}<45\), and trial division by all primes below \(45\), \[ 2,3,5,7,11,13,17,19,23,29,31,37,41,43, \] gives no divisor. Every factor of \(D\) is smaller than \(1979\), so \(1979\nmid D\). Since \(1979\mid pD\), Euclid’s lemma therefore gives \[ \boxed{1979\mid p}. \] This argument does not require \(p/q\) to be in lowest terms.

The solution was checked against John Scholes' IMO solution archive (Kalva).