IMO 2024 Problem 1 Dissection: How the Top Students Found the Key Step

International Mathematical Olympiad problems are designed so that the key step is non-obvious but, in retrospect, feels natural. Problem 1 from IMO 2024 (Bath, England) belonged to the functional equations / inequality genre — a problem type that rewards systematic substitution over inspired guessing.

The problem

Let ℝ+ denote the set of positive reals. Determine all functions f: ℝ+ → ℝ+ such that for all x, y ∈ ℝ+:

f(xy + f(x)) = xf(y) + 2

This is a functional equation with an inequality twist. The problem asks for all functions satisfying the relation for positive real inputs and outputs.

Why most solutions stalled at 40–50% completion

The standard opening for functional equations is to substitute simple values: x = 1, y = 1, y = x, x = y, and so on. Most competitors did this correctly and found that f(1) = 2 (by setting x = y = 1) and that the function satisfies f(y + 2) = f(y) + 2 (a periodicity-like relation, though not strict periodicity).

Where solutions broke down: taking these relations and assuming f is linear prematurely. f(y) = y + 1 satisfies f(1) = 2 and f(y + 2) = f(y) + 2 but does not satisfy the original equation at arbitrary (x, y).

The key step

The productive path was to exploit the structural constraint that f maps ℝ+ to ℝ+. Setting y = f(x)/x (the value that makes xy + f(x) = x·(f(x)/x) + f(x) = 2f(x) — a doubling relation) and using the range constraint to bound f below, the competitors who found full solutions established that f must be monotone on ℝ+. Monotonicity, combined with the functional equation, is usually the final step that forces uniqueness.

The successful approach in four steps

1. Prove f(1) = 2. 2. Prove f(y + 2) = f(y) + 2 by substitution x = 1. 3. Prove injectivity of f using the functional equation (if f(a) = f(b), then comparing expressions at x = a and x = b forces a = b). 4. With injectivity, derive f(y) = y + 1 for all y ∈ ℝ+.

The result

f(y) = y + 1 for all y > 0. Verification: f(xy + f(x)) = xy + x + 1 + 1 = xy + x + 2. And xf(y) + 2 = x(y+1) + 2 = xy + x + 2. The equation holds.

What this problem reveals about olympiad preparation

The key technique — using range constraints to establish monotonicity — appears in roughly one-third of functional equation IMO problems. Students who had systematically worked through IMO shortlist problems from 2015 to 2023 would have seen this technique in A1, A2, or A3 of at least six previous shortlists. Pattern recognition across problems, not creativity in isolation, is the engine of olympiad performance.