I was looking at IMO 2005, problem 2:

Let $a_1, a_2, . ...$ be a sequence of integers with infinitely many positive terms and infinitely many negative terms. Suppose that for each positive integer $n$, the numbers $a_1, a_2, ...a_n$ leave $n$ different remainders on division by $n$. Prove that each integer occurs exactly once in the sequence.

I thought I was more fluent in math English but I was uncertain of the meaning of term “remainder”. I take it that it has the meaning least absolute remainder and e.g. that the remainder of $9$ divided by $6$ is probably not both $-3$ and $3$(?). For $l$ and $m < n$, $a_l≠a_m$ mod ($n$) implies that $a_l≠a_m$. If there also is a one-to-one correspondence between the “remainder” of $a_k$ to $a_k$ mod ($n$) and there are n different terms in the “remainder” series $a_1$ to $a_n$, the problem seems easy, or is it too easy because I got it wrong?

Edit: Slade pointed out that a the proof needs more. Unfortunately my first attempt to prove the missing second part was wrong and had to be deleted - I am still working on this.

Edit2: To avoid the term “remainder” I intended to treat the positive and negative part of the series separately and prove that all integers, positive and negative respectively, must occur once for the two parts and then adding the two series (worrying later about zero possibly occurring twice). However it seems that for e.g. a positive series $a_n=A+n$ the first requirement holds without the numbers $0$ to A being in the series. It seems therefore that there is no way out except to understand the term “remainder”.


You've proved that no integer can occur twice. You also need to prove that every integer occurs somewhere in the sequence, which is much harder.

  • $\begingroup$ could I ask you to look at the edited question with an attempt to prove the difficult part? It is too long to be a comment. $\endgroup$ – Mikael Jensen Oct 14 '18 at 18:53

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