Home > Uncategorized > Balkan Mathematica Olympiad 2017 – Problem 3

Balkan Mathematica Olympiad 2017 – Problem 3


Problem 3. Let {\mathbb{N}} denote the set of positive integers. Find all functions {f:\mathbb{N}\longrightarrow\mathbb{N}} such that

\displaystyle n+f(m)\mid f(n)+nf(m)

for all {m,n\in \mathbb{N}}

Solution: Note that we obviously have

\displaystyle n+f(m) | n^2+nf(m)

and using the hypothesis we obtain that

\displaystyle n+f(m) | f(n)-n^2,

for every {n,m \in \Bbb{N}}. Now we have two options. Suppose the image of {f} is unbounded. Then, since if we fix {n} it would follow that {f(n)-n^2} has arbitrarily large divisor, which is not possible unless {f(n)=n^2}. This is one solution, as it can easily be checked.

The other alternative is that {f(\Bbb{N})} is bounded. If this is true, then there is a value of {f} which is repeated for an infinite increasing sequence {n_k}: {f(n_k) = M}. It follows that

\displaystyle n_k + f(m) | M+n_k f(m)

for every {m} and for {n_k} defined as above. Since the fraction

\displaystyle \frac{M+n_kf(m)}{n_k+f(m)}

is an integer and it converges to {f(m)} as {n_k \rightarrow \infty} it follows that for {n_k} large enough we have {f(m)n_k +f(m)^2 = M+n_kf(m)}. This implies that {f(m)^2 = M}, and this is independent of {m}. Therefore {f} is constant. From the previous relation we have {f(m)^2 = M = f(m)} so the only possible constant is {f\equiv 1}.

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