## Balkan Mathematical Olympiad 2012 Problem 4

Let ${\mathbb{Z}^+}$ be the set of positive integers. Find all functions ${f:\mathbb{Z}^+ \rightarrow\mathbb{Z}^+}$ such that the following conditions both hold:

(i) ${f(n!)=f(n)!}$ for every positive integer ${n}$,

(ii) ${m-n}$ divides ${f(m)-f(n)}$ whenever ${m}$ and ${n}$ are different positive integers.

Balkan Mathematical Olympiad 2012 Problem 4

Solution: First we have $f(k)=f(k)!$ for $k=1,2$, which means that $f(1),f(2) \in \{1,2\}$. We have a few cases:

Case 1: $f(1)=f(2)=1$. Then we know that $n!-1 | f(n!)-1$ and $n!-2 | f(n!)-1$. This means that $f(n!)$ is odd for every $n$ and therefore $f(n)=1$ for every $n \in \Bbb{Z}^+$,

Case 2: $f(1)=f(2)=2$.  First note that $2 | f(3)-2$ so $f(3)$ is even. Then $20 | f(6)-2$ which means that necessarily $f(6)=2$, and furthermore $f(3)=2$. This means that $f(n!)-2 \equiv 2$ modulo $n!-1,\ n!-2,\ n!-3$, and this means that $f(n)!$ is never divisible by $3$. Since $f(n)$ is also even it follows that $f(n)=2$ on $\Bbb{Z}_+$.

Case 3: $f(1)=2,\ f(2)=1$. It follows that $f(6)$ has last digit $7$ and cannot be a factorial.

Case 4: $f(1)=1,\ f(2)=2$. For $n \geq 3$ we have $n!-n | f(n)!-f(n)$ and since the function $n \mapsto n!-n$ is strictly increasing for $n \geq 3$ we find that $f(n) \geq n$, and furthermore $n |f(n)!$, which allows us to deduce from  $n!-n | f(n)!-f(n)$ that $n | f(n)$.

So $f(n)=kn$ with $k \in \Bbb{Z}^+$. Then $k\cdot n!=(kn)!,\ n \geq 3$, which can hold only if $k=1$. In this case we have $f(n)=n,\ \forall n \in \Bbb{Z}^+$.

1. April 30, 2012 at 9:21 am

where is the problem?

• April 30, 2012 at 10:45 am

Do you mean where is the solution? I haven’t posted it yet.

2. May 2, 2012 at 1:07 pm

I didn’t mean that.I’m very sorry Beni because I didn’t read the post carefully.The problem is to find all function f that satisfy the two condition isn’t it?

• May 2, 2012 at 1:29 pm

Yes, that is the problem 🙂