IMO 1979 Problem 1

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 9m29s

Problem

If $p$ and $q$ are natural numbers so that$$ \frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319}, $$prove that $p$ is divisible by $1979$.

Exploration

The alternating sum is examined for small truncations to detect structure.

For $n=1$, the sum is $1$.

For $n=2$, it is $1-\frac12=\frac12$.

For $n=3$, it is $\frac12+\frac13=\frac56$.

For $n=4$, it is $\frac56-\frac14=\frac{7}{12}$.

For $n=5$, it is $\frac{47}{60}$.

These values do not suggest any direct modular periodicity, but they reveal a persistent simplification: the alternating sum consistently collapses into a short harmonic block.

This motivates rewriting the expression by separating odd and even indices rather than attempting modular symmetry on the original alternating form. The earlier approach using a non-preserved involution is avoided because it fails to act on the summation domain.

The expression is checked for structural stability under the transformation $k \mapsto 1979-k$. Direct application to the original index set is invalid, but once the sum is rewritten as a contiguous harmonic block, this symmetry becomes compatible with the summation range.

Problem Understanding

The quantity is

$S = 1 - \frac12 + \frac13 - \cdots + \frac{1}{1319}.$

We write $S = \frac{p}{q}$ in lowest terms and must prove $1979 \mid p$.

The number $1979$ is prime and satisfies

$1979 = 2 \cdot 1319 - 659.$

The goal is to rewrite $S$ into a rational form where multiplication by $1979$ appears explicitly in the numerator while keeping control of cancellation in the denominator so that no factor of $1979$ survives reduction.

Key Observations

The alternating harmonic sum splits into odd and even parts.

The even-index contribution is

$\frac12 + \frac14 + \cdots + \frac{1}{1318} = \frac12\left(1 + \frac12 + \cdots + \frac{1}{659}\right).$

Hence

$S = \left(1 + \frac13 + \cdots + \frac{1}{1319}\right) - \frac12\left(1 + \frac12 + \cdots + \frac{1}{659}\right).$

Grouping terms shows a complete cancellation structure:

$S = H_{1319} - H_{659} = \sum_{k=660}^{1319} \frac{1}{k}.$

The index set $[660,1319]$ has $660$ elements, and the mapping $k \mapsto 1979-k$ preserves this set because

$1979-660 = 1319, \quad 1979-1319 = 660.$

Thus the set is partitioned into $330$ disjoint pairs.

For each pair ${k,1979-k}$,

$\frac{1}{k} + \frac{1}{1979-k} = \frac{1979}{k(1979-k)}.$

This introduces an explicit factor $1979$ at the level of each paired contribution.

Solution

The alternating sum satisfies

$S = 1 - \frac12 + \frac13 - \cdots + \frac{1}{1319}.$

Rewrite by separating odd and even terms:

$S = \sum_{k=1}^{1319} \frac{1}{k} - 2\sum_{k=1}^{659} \frac{1}{2k}.$

The even sum simplifies to

$2\sum_{k=1}^{659} \frac{1}{2k} = \sum_{k=1}^{659} \frac{1}{k},$

hence

$S = \sum_{k=1}^{1319} \frac{1}{k} - \sum_{k=1}^{659} \frac{1}{k} = \sum_{k=660}^{1319} \frac{1}{k}.$

Let

$I = {660,661,\dots,1319}.$

The involution $k \mapsto 1979-k$ maps $I$ onto itself because it swaps endpoints and is strictly decreasing, hence bijective on this interval.

Partition $I$ into disjoint pairs ${k,1979-k}$ with $k \le 1979-k$. There are no fixed points because $k = 1979-k$ would imply $k = \frac{1979}{2}$, not an integer.

Thus,

$S = \sum_{k \in I,; k \le 1979-k} \left(\frac{1}{k} + \frac{1}{1979-k}\right).$

Each pair satisfies

$\frac{1}{k} + \frac{1}{1979-k} = \frac{1979-k + k}{k(1979-k)} = \frac{1979}{k(1979-k)}.$

Therefore,

$S = 1979 \sum_{k \in I,; k \le 1979-k} \frac{1}{k(1979-k)}.$

Let

$T = \sum_{k \in I,; k \le 1979-k} \frac{1}{k(1979-k)}.$

Then $S = 1979T$.

Every denominator $k(1979-k)$ is not divisible by $1979$ since neither factor is a multiple of $1979$, so the reduced form of $T$ has denominator coprime to $1979$. Writing $T = \frac{a}{b}$ in lowest terms with $1979 \nmid b$, we obtain

$S = \frac{1979a}{b}.$

Since $\frac{p}{q}$ is the reduced form of $S$, reduction cannot cancel the factor $1979$ because $1979 \nmid b$, hence the numerator $p$ is divisible by $1979$.

This completes the proof. ∎

Verification of Key Steps

The identity

$S = \sum_{k=660}^{1319} \frac{1}{k}$

matches direct algebraic simplification for alternating harmonic sums and holds without exception in the tested small cases.

The mapping $k \mapsto 1979-k$ preserves the interval $[660,1319]$ because it reverses endpoints and is strictly monotone, giving a bijection on a finite set.

Each paired sum computation

$\frac{1}{k} + \frac{1}{1979-k} = \frac{1979}{k(1979-k)}$

is exact arithmetic with common denominator $k(1979-k)$ and introduces no hidden assumptions.

No denominator in the construction contains a factor $1979$, since $k < 1979$ and $1979-k < 1979$ for all $k \in I$, so no cancellation of the factor $1979$ can occur during reduction of the final fraction.

Alternative Approaches

The same result can be obtained by writing the sum as

$S = \sum_{k=660}^{1319} \frac{1}{k} = \frac{1}{1979} \sum_{k=660}^{1319} \frac{1979}{k}$

and then applying a direct pairing argument at the level of residues modulo $1979$, where the interval is symmetric under inversion around $\frac{1979}{2}$.

Another approach uses the identity $1979 = 2\cdot1319 - 659$ to convert the sum into a symmetric harmonic block and then applies the involution $k \mapsto 1979-k$ to induce factorization of the numerator.