# Number Theory Problem

Determine all positive integers $n \geq 2$ such that, given any integers $a_1, a_2, ... , a_n$ , it is possible to find integers i and j ( $1 \leq i < j \leq n$ ) such that:
$a_i + a_{i+1}$ and $a_j + a_{j+1}$ leave the same remainder in the division by $n$
Note: Consider $a_{n+1} = a_1$.

Sorry, I can't really write in $\LaTeX$.

Note by John Smith
3 years, 6 months ago

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Re Latex, you've got the code. You just needed to add in the Latex brackets of $\backslash ( \quad \backslash)$. I've edited your note so you can see my edits.

Staff - 3 years, 6 months ago

@Calvin Lin Thank you very much! And do you have any idea of how I can proceed in this problem? I haven't made any progress.

- 3 years, 6 months ago

What have you tried? Where are you stuck?

What type of problem does this resemble? What approaches do you think should be pursued?

Staff - 3 years, 6 months ago

- 3 years, 6 months ago

I could do it if the integers didn't have to be onsecutive in the set of integers given, because there are n numbers with n possible remainders for them, so it would be easy. But like this, I really don't know how to start.

- 3 years, 6 months ago

That's a good start. You realized that if we set $b_i = a_i + a_{i+1}$, we would get $n$ numbers with $n$ possible remainders.

1. Suppose that all of the remainders are different. What conclusion can we draw?
2. What is a necessary condition for the remainders to not be different? Let's try small cases.
3. What is a necessary condition for the remainders to be different?
4. For each value that satisfies the necessary condition, can we create a sequence that gives us those remainder? Let's use really wishful thinking even.
5. Does this help us classify all positive integers?

Staff - 3 years, 6 months ago

@Calvin Lin I think that if all sums had different remainders, the remainders of the numbers in odd positions of the set given would all be different and the ones in even positions would all be different.

- 3 years, 6 months ago

That's a thought. Continue developing it. Try small cases. Play around.

For $n = 2, 3, 4 , \ldots$, we have at most $n^2$ sets of remainders to test, so you should be able to make some initial guesses pretty quickly.

Staff - 3 years, 6 months ago

@Calvin Lin I couldn't do it. I did what you said and tried a lot for $n=2 , 3, 4 ...$ but I couldn't make any progress. I should start with easier problems maybe. Have you solved this? How?

- 3 years, 6 months ago

I've not worked it out completely, but I believe the way I outlined below leads to a solution. I do believe that looking at small cases will help you get towards the approach, especially if you think of generalizing (instead of proving specific cases)

What conclusions did you get for small values? Is there a pattern?

This problem seems typical of pigeonhole principle, which is what you were suggesting when you said "n numbers, n possible remainders". We want to get some kind of conclusion / conclusion when all the remainders are distinct. For example, one thing that pops into mind, is the sum of all these remainders.

Staff - 3 years, 6 months ago