# 2004 IMO Shortlist Problems/A2

## Problem

(*Mihai Bălună, Romania*)
An infinite sequence of real numbers satisfies the condition

for every ,

with and positive and distinct. Can this sequence be bounded?

*This was also Problem 4 of the 2005 German Pre-TST and Problem 1 of the first 2005 black MOP test. It was Problem 3, Day 2 of the 2005 Romanian TST in the following form:*

A sequence of real numbers is called a *bs* sequence if , for all . Prove that a bs sequence is bounded if and only if the function given by , for all is the null function.

## Solution

### Solution 1

We note that each of the must be nonnegative.

**Lemma 1.** If the two initial terms of the sequence are nonzero and distinct, then every term of the sequence is nonzero and no two consecutive terms are equal.

*Proof.* We proceed by induction; we are given a base case. If , then , so . Furthermore, since , .

Consider a sequence of positive reals which obeys the recursive relation

**Lemma 2.** .

*Proof.* We note that if , then and . Thus if , we must have , a contradiction.

**Lemma 3.** Let , . Then .

*Proof.* For , easy induction yields , , .

Now, if is even, we have , which is less than by the definition of , and , , . Since and add to , one of them must be at most , and the lemma follows.

On the other hand, if is odd, then set and we have , , and , so as before, at least one of , must be at most . In fact, we have , so the minimum of is at most , upholding the lemma, since in this case .

Now, suppose that is bounded, i.e., there exists some real greater than each . By setting equal to , we generate the first of the in reverse, and from Lemma 2, we can see that is a lower bound of the . But by making greater than , by applying Lemma 3, we obtain the result , which is a contradiction when are distinct and greater than 0, by Lemma 1.

Now, if for all natural , all the nonzero must be equal and the sequence is bounded. On the other hand, if always holds, then either and , or one of is equal to zero and is equal to the other one; hence by induction, all the nonzero terms of the sequence are equal, and is always equal to zero. Hence if for some , , then there exist two distinct positive consecutive terms of sequence, which is then unbounded as proven above.

### Solution 2

The given recursive condition is equivalent to if , and otherwise. Note that if , then .

**Lemma 1: ** Let be the minimal element of . Then .

*Proof: * Suppose is the minimal value of such that . Then , so , and . This contradicts our assumption of minimality.

Let ; we claim that , which would imply that is unbounded, in turn implying that is unbounded.

- Suppose ; then or . In the former case, so , and . In the latter case, .
- Suppose , ; then , and . Then .

Then , so is unbounded, as desired.

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*

## See also

**Comment by the proposer.** The following statements are equivalent for a sequence of real numbers satisfying for :

i) the sequence is bounded;

ii) the function defined by , for , is identically zero;

iii) the sequence is of the form with .