RIGOR / MORTISPUBLIC OLYMPIAD PROOF CHALLENGES · NO SIGNUP
PUBLIC PROOF TRAP · NO. 01 / POLYNOMIAL DYNAMICS

The theorem is true.
The proof is not.

Can you locate the first mathematical gap in an argument that nevertheless reaches the correct answer? Inspired by IMO 2006 Problem 5.

NUMBER THEORYPOLYNOMIALSERROR LOCALIZATION
01 / CHALLENGE

Where does the argument break?

Theorem. Let P ∈ ℤ[x] have degree d ≥ 2. For a positive integer k, let S = {t ∈ ℤ : P∘k(t) = t}. Then |S| ≤ d.
  1. The set S is finite and P(S) ⊆ S. Indeed, P∘k(x) − x has degree dk, while P∘k(P(t)) = P(P∘k(t)).
  2. For every distinct u,v ∈ S, integer polynomial divisibility around the two periodic orbits gives |P(u) − P(v)| = |u − v|.
  3. For a = min S and b = max S when |S| ≥ 2, distance preservation implies {P(a),P(b)} = {a,b}.
  4. Thus P(t) = t for all t ∈ S. Hence P(x) − x has at least |S| distinct roots; its degree is d, proving the claim.
YOUR THREE TASKS
  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 A COMPLETE REFERENCE REPAIR (SPOILERS) ▾

The missing reflection branch

First unsupported inference: Step 4. A distance-preserving map of a finite ordered subset of the real line may be a reflection rather than the identity.

Counterexample: choose P(x) = (x−1)2 and k=2. Then P(0)=1 and P(1)=0, so both 0 and 1 lie in S, yet neither is a fixed point of P.

Why Step 2 is justified: define δᵢ = P∘i(u) − P∘i(v). For distinct u,v in S, all these differences are nonzero, and δ₀ | δ₁ | ··· | δₖ = δ₀. Taking absolute values proves the claimed equality.

The repair: if |S| ≤ 1, we are done. Otherwise set a = min S, b = max S. The extreme images are either (a,b) or (b,a).

In the first case, P(a)=a and for every t in S we have |P(t)−a|=|t−a|=t−a. Because P(t) ∈ [a,b], it follows that P(t)=t. Every point of S is therefore a root of P(x)−x.

In the second case, P(a)=b, so |P(t)−b|=|t−a|=t−a. Since P(t) ∈ [a,b], this gives P(t)=a+b−t. Every point of S is a root of P(x)+x−(a+b).

Both resulting polynomials are nonzero and have degree d, because d≥2. Hence |S|≤d in either case. Q.E.D.

Adapted from IMO 2006 Problem 5. The theorem is classical; the deliberately flawed presentation and accompanying critique are practice material, not a research novelty claim.

What this is not: 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

Explore the proof-forensics laboratory.

111 proof audits

Read Olympiad-inspired fragments, identify the earliest unsupported step and write a mathematical repair.

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Six selected full proof fragments with neutral titles, local reviewer notes and optional JSON export.

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