RIGOR / MORTISPUBLIC OLYMPIAD PROOF CHALLENGES · NO SIGNUP
PUBLIC PROOF TRAP · NO. 02 / POLYNOMIAL DYNAMICS

One lemma is false.
The theorem survives.

Can you locate the first mathematical gap in an argument that nevertheless reaches the correct answer? Inspired by IMO 2021 Problem 6.

COMBINATORICSCOUNTINGFALSE INJECTIVITY
01 / CHALLENGE

Where does the argument break?

Theorem. Let m ≥ 2, and let A={a₁,…,aₙ} be a finite integer set. Suppose each of m,m²,…,mm is the sum of some subset of A. Prove n ≥ m/2.
  1. For j=1,…,m choose a subset Bⱼ of A with sum mʲ. The mm weighted sums N=t₁m+t₂m²+⋯+tₘmm (0≤tⱼ<m) are all distinct by uniqueness of base-m digits.
  2. Every N equals ∑ᵢcᵢaᵢ, where cᵢ=∑ⱼtⱼ·1[aᵢ∈Bⱼ]. Thus 0≤cᵢ≤m(m−1)<m² for every i.
  3. Distinct integer coefficient vectors (c₁,…,cₙ) always yield distinct sums ∑ᵢcᵢaᵢ.
  4. There are at most m2n coefficient vectors. Hence mm≤m2n, and therefore n≥m/2.
YOUR THREE TASKS
  1. Find the first false assertion.
  2. Give two distinct allowed coefficient vectors with the same sum.
  3. Recover the sharp n≥m/2 counting bound without the false assertion.
  1. Locate the first unjustified inference, without merely saying the proof is incomplete.
  2. Produce an explicit integer polynomial of degree at least 2 that invalidates that inference.
  3. Replace the broken step with a complete proof of the theorem.
REVEAL THE REFERENCE COUNTEREXAMPLE & COMPLETE REPAIR ▾

The injection we never needed

First false step: 3. For m=2, take A={2,4}, where 2 and 4 are subset sums as required. The distinct allowed coefficient vectors (2,0) and (0,1) both yield 4.

Repair: We only need every target N to be represented by at least one allowed vector. Two different target integers cannot be produced by the same coefficient vector. Therefore the mm distinct target integers require at least mm different vectors, regardless of any collisions elsewhere.

For each coordinate 0≤cᵢ≤m(m−1), giving exactly m(m−1)+1≤m² possible integers. There are at most m2n allowed vectors. Combining gives mm≤m2n, hence n≥m/2. Q.E.D.

The correct relation is a surjection from the allowed vectors onto the constructed target set—not an injection from every vector to its value.

Adapted from IMO 2021 Problem 6. This is a public, author-labelled practice example and not a blinded research item.

Public practice only: An undisclosed test of human participants. This challenge is fully public; its reference repair is intentionally available. No identifying information, study answers or analytics are collected by this page.
02 / MORE MATHEMATICS

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