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Balti Tee 2011 · Valikvooru ülesanne

Algebra

Find all functions ff from the set of all positive integers to the same set such that, for all positive integers a1,…,aka_1, \dots, a_k with k>0k > 0, the sum a1+⋯+aka_1 + \dots + a_k divides the sum f(a1)+⋯+f(ak)f(a_1) + \dots + f(a_k).

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Answer: All functions given by f(n)=anf(n) = an, a∈Na \in \mathbb{N}.

Suppose that ff is a function that satisfies the conditions of the problem. We claim that f(n)=f(n−1)+f(1)f(n) = f(n-1) + f(1) for all integers n>1n > 1. Indeed, for any integer m>nm > n, we have m∣f(n)+f(m−n)m \mid f(n) + f(m-n) and m∣f(n−1)+f(1)+f(m−n)m \mid f(n-1) + f(1) + f(m-n) by conditions of the problem. Hence the difference f(n)−(f(n−1)+f(1))f(n) - (f(n-1) + f(1)) is also divisible by mm. As mm was arbitrary, this implies that f(n)−(f(n−1)+f(1))f(n) - (f(n-1) + f(1)) is divisible by an infinite number of different integers, i.e., is equal to 00. This completes the proof of the claim.

Easy induction now gives that necessarily f(n)=nf(1)f(n) = n f(1). It remains to verify that all functions of the form f(n)=anf(n) = a n satisfy the conditions of the problem, which is straightforward.