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Algebra Difficulty 8.3 Shortlist Prove it Romania

Define a sequence of integers by a0=1a_0 = 1, and an=k=0n1(nk)aka_n = \sum_{k=0}^{n-1} \binom{n}{k} a_k, n1n \ge 1. Let mm be a positive integer, let pp be a prime, and let qq and rr be non-negative integers. Prove that the difference apmq+rapm1q+ra_{p^m q+r} - a_{p^{m-1} q+r} is divisible by pmp^m.

Solution

Consider the R\mathbb{R}-vector space R[X]\mathbb{R}[X] of all polynomials with real coefficients and define an R\mathbb{R}-linear functional L:R[X]RL: \mathbb{R}[X] \to \mathbb{R} by LXn=anLX^n = a_n, n=0,1,2,n = 0, 1, 2, \dots. Thus, if f=kαkXkf = \sum_k \alpha_k X^k, then Lf=kαkakLf = \sum_k \alpha_k a_k. Since (X+1)n=k=0n(nk)Xk(X+1)^n = \sum_{k=0}^n \binom{n}{k} X^k, n1n \ge 1,
L(X+1)n=k=0n(nk)LXk=k=0n(nk)ak=k=0n1(nk)ak+an=2an=2LXn, L(X+1)^n = \sum_{k=0}^{n} \binom{n}{k} LX^k = \sum_{k=0}^{n} \binom{n}{k} a_k = \sum_{k=0}^{n-1} \binom{n}{k} a_k + a_n = 2a_n = 2LX^n,
so Lf(X+1)=2Lf(X)f(0)Lf(X+1) = 2Lf(X) - f(0) for every polynomial ff in R[X]\mathbb{R}[X]. In particular, take f=(Xk)f = \binom{X}{k} and use the relation (X+1k)=(Xk)+(Xk1)\binom{X+1}{k} = \binom{X}{k} + \binom{X}{k-1}, k1k \ge 1, to get L(Xk)=L(Xk1)L\binom{X}{k} = L\binom{X}{k-1}, k1k \ge 1, and deduce that L(Xk)=1L\binom{X}{k} = 1, k=0,1,2,k = 0, 1, 2, \dots. Further, if a polynomial ff in R[X]\mathbb{R}[X] is integral valued, i.e., f(k)f(k) is integral for every integral kk, then f=kαk(Xk)f = \sum_k \alpha_k \binom{X}{k} for some integers αk\alpha_k, so Lf=kαkLf = \sum_k \alpha_k is an integer. Finally, since apmapm1(modpm)a^{p^m} \equiv a^{p^{m-1}} \pmod{p^m} for all integers aa,
f=pm(Xpmq+rXpm1q+r) f = p^{-m} \left( X^{p^m q+r} - X^{p^{m-1} q+r} \right)
is an integral valued polynomial in R[X]\mathbb{R}[X], so Lf=(apmq+rapm1q+r)/pmLf = (a_{p^m q+r} - a_{p^{m-1} q+r})/p^m is an integer, as required.

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