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Number theory Difficulty 6.2 National olympiad Prove it

Let (an)nN\left(a_{n}\right)_{n \in \mathbb{N}^{*}} be a sequence of integers. We say that (an)nN\left(a_{n}\right)_{n \in \mathbb{N}^{*}} is a permutation of N\mathbb{N}^{*} if for every mNm \in \mathbb{N}^{*}, there exists a unique nNn \in \mathbb{N}^{*} such that an=ma_{n}=m. Show that there exists a sequence (Pi)iN=\left(P_{i}\right)_{i \in \mathbb{N}^{*}}= ((ai,j)jN)iN\left(\left(a_{i, j}\right)_{j \in \mathbb{N}^{*}}\right)_{i \in \mathbb{N}^{*}} of permutations of N\mathbb{N}^{*} satisfying for all kNk \in \mathbb{N}^{*} and all 1i1<i2:1 \leqslant i_{1}<i_{2}:

Sk(Pi1)Sk(Pi2) S_{k}\left(P_{i_{1}}\right) \mid S_{k}\left(P_{i_{2}}\right)

where Sk(Pi)S_{k}\left(P_{i}\right) denotes the sum of the first kk elements of Pi:Sk(Pi)=j=1kai,jP_{i}: S_{k}\left(P_{i}\right)=\sum_{j=1}^{k} a_{i, j}.

## Solved by Matthieu Vogel

Solution

solved by Matthieu Vogel

Let u:NNu: \mathbb{N} \rightarrow \mathbb{N}^{*} be a sequence of non-zero natural numbers. We say that an index ii of this sequence is nice if u1++uiu_{1}+\cdots+u_{i} and u1++ui+1u_{1}+\cdots+u_{i+1} are coprime.

Since divisibility is transitive, it is necessary and sufficient to find P:iNP: i \rightarrow \mathbb{N}^{*} such that Sk(Pi)Sk(Pi+1)S_{k}\left(P_{i}\right) \mid S_{k}\left(P_{i+1}\right) for all i,ki, k. We show that there exists a permutation of N\mathbb{N}^{*} that is nice and that for any nice permutation PiP_{i}, we can construct a nice permutation Pi+1P_{i+1} such that Sk(Pi)Sk(Pi+1)S_{k}\left(P_{i}\right) \mid S_{k}\left(P_{i+1}\right) for all i,ki, k.

First, let's find a permutation a1,a2,a_{1}, a_{2}, \ldots of N\mathbb{N}^{*} that is nice. We define it by induction. a1=1a_{1}=1. If ii is even, we take ai+1=pa_{i+1}=p a prime a1,,ai\neq a_{1}, \ldots, a_{i} large enough so that gcd(a1++ai,ai++ai+ai+1)=gcd(a1++ai,p)=1\operatorname{gcd}\left(a_{1}+\cdots+a_{i}, a_{i}+\cdots+a_{i}+a_{i+1}\right)=\operatorname{gcd}\left(a_{1}+\cdots+a_{i}, p\right)=1, hence ii is a nice index. If iii \geqslant i is odd, we take ai+1=min(N\{a1,,ai})a_{i+1}=\min \left(\mathbb{N}^{*} \backslash\left\{a_{1}, \ldots, a_{i}\right\}\right), hence the sequence is a permutation of N\mathbb{N}^{*}.

Secondly, let a1,a2,a_{1}, a_{2}, \ldots be a nice sequence. Construct a sequence b1,b2,b_{1}, b_{2}, \ldots that is nice such that a1++aib1++bia_{1}+\cdots+a_{i} \mid b_{1}+\cdots+b_{i} for all ii. We construct the sequence bb by induction. We take b1=a1b_{1}=a_{1}. Then, to construct bib_{i}, if ii is not nice in the sequence aa, we take bib_{i} such that b1++bi0(moda1++ai)b_{1}+\cdots+b_{i} \equiv 0\left(\bmod a_{1}+\cdots+a_{i}\right) and bib1,,bi1b_{i} \neq b_{1}, \ldots, b_{i-1}. If ii is a nice index in aa, we will construct both bib_{i} and bi+1b_{i+1}. Let A=a1++aiA=a_{1}+\cdots+a_{i} and A=a1++ai+1A^{\prime}=a_{1}+\cdots+a_{i+1}, AA and AA^{\prime} are coprime because ii is a nice index. We will show that if xb1,b2,ai1x \neq b_{1}, b_{2}, \ldots a_{i-1}, we can find bi,bi+1b_{i}, b_{i+1} such that all divisibility conditions are satisfied and bi+1=xb_{i+1}=x. It will then suffice to take x=min(N\{b1,,bi1)x=\min \left(\mathbb{N}^{*} \backslash\left\{b_{1}, \ldots, b_{i-1}\right)\right. an infinite number of times so that the sequence bb is a permutation of N\mathbb{N}^{*} and xx a large enough prime so that ii is a nice index in the sequence bb an infinite number of times so that the sequence bb is nice. If we fix bi+1=xb_{i+1}=x, it is sufficient to find bib_{i} such that b1++bi0(modA)b_{1}+\cdots+b_{i} \equiv 0(\bmod A) and b1++bi+x0(modA)b_{1}+\cdots+b_{i}+x \equiv 0\left(\bmod A^{\prime}\right). It exists by the Chinese Remainder Theorem and we take it large enough so that it is different from a1,,ai1,xa_{1}, \ldots, a_{i-1}, x.

## VIII. Memorable Quotes

- Tristan: "I'm studying math, I haven't seen numbers in years."
- Rémi: "No, you can't be in two places at the same time. Physicists can, but not you."

Tristan: "That's rather positive, it means you're not a physicist."

- Aurélien: "If you choose fascist friends, it will show."

— Pierre-Marie: "I don't say stupid things, ask Aurélien."

- Zinedine: "Especially Paul, he's a human."

— Théo, talking about graphs: "This tree is ugly, we want to cut it down."

- Théo: "So there's something Tristan realizes. Normally he's already realized it. But well! It's Tristan."

- Pierre-Marie: "We have a pretty little equation that's wandering in the forest."

- Pierre-Marie: "Anyone who uses Zsigmondy should jump out the window." Later in the proof, "Well, we're going to have to use Zsigmondy."

- Rémi: "The people who got 7/7 can be counted on the fingers of a one-armed man."

- Anatole, 13 years old, to first-year students: "Good night, little ones!"

- Gaspard: "Tic tac toc without pif paf pof is much better than tic tac."

— "I have to write a letter to someone."

Martin: "Is her name Élise, just in case?"

- Rémi: "I'm a pretty primitive person."

- "When will we get the T-shirts?" Rémi: "Yes."

The T-shirts only arrived on the last day.

- Georges: "The anims are all young and dynamic, even François Lo Jacomo is young and dynamic!"

— Martin: "There's fa, there's do. What else is there as a note already?"

- Alexander: "Everything is valued." Pierre-Marie: "Nothing is serious."

— Victor: "A 4x3 square."

- Victor: "There are bald people with zero hair."

- Théo: "So yes, I write very badly... If you can't decipher me, it's your fault."

- At the departure of the anims:

"Goodbye!"

"I don't think they can hear us."

"I can shout!"

"Even if you shout, I don't count on you to bring them back."

- Emile: "For the class, I take a break at 4:30 PM." The classes end at 4:30 PM.

- Eva: "It would be great to have a compass in the eye."

- Théo: "You're confusing Euclidean division with Euclidean division."

- Emma: "I've never completed anything in my life."

- Matthieu Vogel: "Do you often go to Bodo's camp?"

Amélie: "Yes, every time."

Matthieu V.: "That's why I'm bad at math."

- Théo, during the anims meeting: "This morning, I gave a sleeping class to the group."

- Stéphane: "Stanislas is modern, but in the old style."

- Emma: "I see the world in color, and it hurts my head."

- Théo: "But yes, it's possible with the kiwi card."

— Raphaël: "There's a shooting going on downstairs."

Everyone looks towards the window.

Raphaël, disappointed: "Okay... you can go look for two minutes." Everyone rushes to the window.

[^0]: 1. We will abuse the terminology and refer to the remainder of the division by 7 as the value modulo 7.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.