IMO 2009 N5 #
Let $S$ be a set and $T : S → S$ be a function. Prove that there does not exist a non-constant polynomial $P ∈ ℤ[X]$ such that for every positive integer $n$, $$ \#\{x ∈ ℤ : T^n(x) = x\} = P(n). $$
Solution #
We follow a variant of Solution 2 of the official solution. Namely, $p$ does not have to be prime; we have $P(0) ≡ P(n) \pmod{q}$ for any $n > 0$.
Extra lemma #
If a₀ is T-periodic, then a has the same T-orbit as a₀
if and only if a is in the T-orbit of a₀.
Start of the problem #
We say that a function T : α → α is nice if T^n has
finitely many fixed points for every positive integer n.
- toFun : α → α
Fixed points of $T^n$, except we set the set to be empty at $0$ for convenience.
- nth_fixed_pts_spec' (n : ℕ) (hn : n > 0) (a : α) : a ∈ self.nth_fixed_pts n ↔ Function.IsPeriodicPt self.toFun n a
Instances For
Equations
- IMOSL.IMO2009N5.NiceFun.instFunLike = { coe := IMOSL.IMO2009N5.NiceFun.toFun, coe_injective' := ⋯ }
nth_fixed_pts n consists of points a such that T^n(a) = a.
The finite set of points of T-period exactly n.
Equations
- T.period_n_pts n = {a ∈ T.nth_fixed_pts n | Function.minimalPeriod (⇑T) a = n}
Instances For
period_n_pts 0 = ∅.
The finite set period_n_pts is indeed the set of T-period n points if n > 0.
An alternate version of period_n_pts_spec that, in addition,
says that there are no "T-period 0 points".
Any member of period_n_pts is a member of periodicPts T.
If T.period_n_pts n has an element, then n > 0.
If a has T-period n, then T^k(a) has T-period n.
Summation formula for the number of fixed points #
T^n(a) = a if and only if a has T-period dividing n.
Note: Due to the formulation of nth_fixed_pts and Nat.divisors,
the implemented formulation works even if n = 0.
The sets of points with T-period n are pairwise disjoint across all n.
The set of nth fixed points is the union, over all d ∣ n,
of the set of elements of T-period d.
Summation formula for the number of points a such that T^n(a) = a.
Note: This is one of the key observations used in the solution.
The number of points of period n is divisible by n. #
The "periodic orbit" of the points of T-period n has size n.
The size of period_n_pts is n times the number of corresponding orbits.
The set of points with T-period n has size divisible by n.
Note: This is one of the key observations used in the solution.
For any n and for any prime q, the number of fixed points
of T^n and T^{nq} are congruent modulo q.
Final solution