Assume $p<q$ be the smallest two prime divisors of $n$. Then $d_{k-1}=\frac{n}{p}$. Assume $d_m=\frac{n}{q}$ for some $m$, and $d_{m+1}=\frac{n}{p^{c+1}}$ and $d_{m+2}=\frac{n}{p^c}$ for some nonnegative integer $c$. Then, since $d_m\mid d_{m+1}+d_{m+2}$, $p^{c+1}\mid q+pq$ and $p|q$ which is a contradiction. Therefore $n$ does not have two distinct prime factors; $n$=$p^t$ for some prime $p$ and a positive integer $t\neq1$. It's easy to show that this suffices.

From $d_{k-2} \mid d_{k-1} + d_{k}$ one can get that $d_{k-2} \mid d_{k-1}$ since $d_{k-2} \mid n = d_k$ Then $d_2 = \frac{n}{d_{k-1}} \mid \frac{n}{d_{k-2}} = d_3$ so $d_2 \mid d_3$. Here, since $d_2 \mid d_3 + d_4$, we get that $d_2 \mid d_4$. Again, we can easily show $d_{k-3} | d_{k-1}$ in the same way, which leads to $d_{k-3} \mid d_{k-2}$ By induction, $d_1 \mid d_2 \mid \cdots \mid d_k$ If $p \mid n$ and $q \mid n$ for prime $p > q$ $\frac{n}{p} \mid \frac{n}{q}$ so $q \mid p$ and this is contradiction Hence, $n = p^{k-1}$ This indeed fits the condition

The answer is prime powers, which obviously work. Note that the three largest divisors of $n$ are either $\left \{ \tfrac{n}{q}, \tfrac{n}{p}, n \right \}$ for distinct prime divisors $p$ and $q$ of $n$, or $\left \{ \tfrac{n}{p^2}, \tfrac{n}{p}, n \right \}$ for some prime divisor $p$ of $n$. In the former case we have a contradiction, since \[\frac{\frac{n}{p} + n}{\frac{n}{q}} = \frac{q(p+1)}{p},\]obviously not an integer. So, the three largest divisors of $n$ are $\frac{n}{p^2}, \frac{n}{p}$ and $n$. With similar reasoning, now consider the fourth largest divisor. It is either $\tfrac{n}{q}$ or $\tfrac{n}{p^3}$; the former option fails, since \[ \frac{\frac{n}{p} + \frac{n}{p^2}}{\frac{n}{q}} = \frac{q(p+1)}{p^2}, \]obviously not an integer either. We repeat this reasoning to deduce that $n$ is just $p^{k-1}$.

HUH how is this IMO level?

The answer is $p^e$ for all primes $p$ and $e>1$, which obviously works. Let $p$ be the smallest prime factor of $n$, we claim that $p$ is the only prime factor of $n$. We use induction downward to prove $d_i = \frac{n}{p^{k-i}}$. The base case is obvious. For the inductive step, if we instead had a different prime $q$, then \[ \frac{n}{q} \mid \frac{n}{p^{k-i}} + \frac{n}{p^{k-i-1}}, \]but this is obviously false. We are done.

This NT seems familiar. Maybe it's an old problem?

LoloChen wrote: This NT seems familiar. Maybe it's an old problem? The only related problem I can think of is IMO 2002 P4, but it's not even close.

LoloChen wrote: This NT seems familiar. Maybe it's an old problem? Maybe JBMO 2002 is similar with this problem but this is even easier than that https://artofproblemsolving.com/community/c6h58610p358021

ihatemath123 wrote: The answer is prime powers, which obviously work. Note that the three largest divisors of $n$ are either $\left \{ \frac{n}{q}, \frac{n}{p}, n \right \}$ for distinct prime divisors $p$ and $q$ of $n$, or $\left \{ \frac{n}{p^2}, \frac{n}{p}, n \right \}$ for some prime divisor $p$ of $n$. In the former case we have a contradiction, since \[\frac{\frac{n}{p} + n}{\frac{n}{q}} = \frac{q(p+1)}{p},\]obviously not an integer. So, the three largest divisors of $n$ are $\frac{n}{p^2}, \frac{n}{p}$ and $n$. With similar reasoning, now consider the fourth largest divisor. It is either $\frac{n}{q}$ or $\frac{n}{p^3}$; the former option fails, since \[ \frac{\frac{n}{p} + \frac{n}{p^2}}{\frac{n}{q}} = \frac{q(p+1)}{p^2}, \]obviously not an integer either. We repeat this reasoning to deduce that $n$ is just $p^k$. I think you have a very small mistake isn't it $p^{k-1}$??

I'll only show the more technical part below.

For any $n$, the 3 largest divisors are $n,\frac{n}{p},\frac{n}{q}$ for primes $p<q$ or $n,\frac{n}{p},\frac{n}{p^2}$. However, first case doesn't work because $\frac{\frac{n}{p}+n}{\frac{n}{q}}=\frac{q}{p}+q$ which is not an integer. Repeat for further cases. All should not work for $q$ as you will get $\frac{q(p+1)}{p^k}$ which are not integers. Thus only $p^{k-1}$ work well there goes 10M N prediction more like 0M now

This problem was proposed by me, Santiago Rodriguez from Colombia. I hope that you enjoy it.

Far too easy for the IMO. It's just prime powers, which trivially work. Suppose $n$ isn't a prime power. Clearly the list of divisors goes $\left(1,\cdots,\frac{n}{q}, \frac{n}{p^k},\cdots,\frac{n}{p},n\right)$ for some primes $p,q$ and some $k\in\mathbb{N}$. Thus $\frac{n}{q}\mid \frac{n}{p^k}+\frac{n}{p^{k-1}}$ and hence $$\frac{q(p+1)}{p^{k-1}}\in\mathbb{Z}$$Which is impossible unless $k=1$, and for $k=1$ we need $\frac{n+n/p}{n/q}$ to be an integer, which is impossible.

yofro wrote: Far too easy for the IMO. It's just prime powers, which trivially work. Suppose $n$ isn't a prime power. Clearly the list of divisors goes $\left(1,\cdots,\frac{n}{q}, \frac{n}{p^k},\cdots,\frac{n}{p},n\right)$ for some primes $p,q$ and some $k\in\mathbb{N}$. Thus $\frac{n}{q}\mid \frac{n}{p^k}+\frac{n}{p^{k-1}}$ and hence $$\frac{q(p+1)}{p^{k-1}}\in\mathbb{Z}$$Which is impossible unless $k=1$, and for $k=1$ we need $\frac{n+n/p}{n/q}$ to be an integer, which is impossible. why $\frac{n}{q} < \frac{n}{p^k}$?

Solution : Observe that we have $$d_i | d_{i+1}+d_{i+2} , 1 \leq i \leq k-2$$, then put $i=k-2$ then $$d_{k-2} | d_{k-1}+d_{k}=d_{k-1}+n$$but $$d_{k-2} |n \Longrightarrow d_{k-2} | d_{k-1} .$$Again putting $i=k-3$ we get $$d_{k-3} | d_{k-2}+d_{k-1}$$but on other hand $$d_2=\frac{n}{d_{k-1}} | \frac{n}{d_{k-2}} =d_3 , d_2|d_3+d_4 \Longrightarrow d_2|d_4.$$So $$d_{k-3}|d_{k-1} \Longrightarrow d_{k-3}|d_{k-2} .$$Now using Induction we get $$d_1|d_2|d_3.......d_k$$. Hence $n=p^{k-1}$ is the solution.

khan.academy wrote: why $\frac{n}{q} < \frac{n}{p^k}$? I think that the poster meant $k$ is just an natural number not $\tau{n}$ Then choose $a$ such that $d_1 = 1, d_2 = p, d_3 = p^2, \cdots, d_a = p^{a-1}, d_{a+1} = q$ Here, $\frac{n}{q} < \frac{n}{p^a}$

I want to post a solution as short as I can. Tell me if I made a mistake. Really too easy to be an IMO problem. Quote: $n=p^a$ triviral. If $ n$ has 2 minimal prime divisors $d_1=p<d_t=q$ and $d_{t-1}=p^m$, $\frac n q \mid \frac n {p^m}+\frac n {p^{m-1}}$. $v_p(LHS)>v_p(RHS)$, which leads to contradiction.

$d_{n-2}$ divides $d_{n-1} + d_n$ so $d_{n-2}$ divides $d_{n-1}$ $d_{n-3}$ divides $d_{n-2} + d_{n-1}$ so $d_{n-3}$ divides $d_{n-2}$ repeat to get that $d_a$ divides $d_{a+1}$ where $a$ is random integer So $d_a$ divides $d_{a+1}$, which divides $d_{a+2}$, ...... so $d_a$ divides $d_{a+x}$ where $x$ is integer $> 0$ and anything divides or is divided by anything so $n$ has to be a prime power because if it was $a*b$ where $a$ and $b$ or primes than there will be $a$ and $b$ but $a$ does not divide $b$

Last year, I wrote a terribly wrong solution and never bother to fix it A year later, I finally did it. We claim all numbers that work are of the form $p^n.$ These obviously work as $p^k|p^k+1(1+p)$ for all integer $k.$ Now assume FTSOC there is a number with multiple prime factors. Let the two smallest be $p$ and $q.$ (Edit: with p<q) Let $c$ be the largest integer such that $p^c<q.$ Then there will be three numbers in a row that are $$\frac{N}{q}, \frac{N}{p^c}, \frac{N}{p^c-1}.$$By the problem statement, we can write that $$\frac{\frac{N}{p^c}(1+p)}{\frac{N}{q}}$$is an integer, which simplifies to $(1+p)\frac{q}{p^c}$. Since $gcd(p+1,p)=1,$ and $gcd(q,p)=1,$ the fraction cannot be further simplified and since $p$ is a prime the fraction is obviously not an integer. Therefore, there are no numbers with multiple prime factors that work. Q.E.D.

Sol:- Clearly $p^k$ where $p$ is a prime number (and $k>1$) works. Now suppose a number is divisible by atleast $2$ primes, we let $p<q$ be its smallest $2$ prime factors. Let $r$ be the greatest number such that $p^r<q$. Then $\frac{n}{q},\frac{n}{p^r},\frac{n}{p^{r-1}}$ are $3$ divisors of $n$ in consecutive order. Thus $\frac{n}{q}|(\frac{n}{p^r}+\frac{n}{p^{r-1}}) \implies p^r|q(p+1)$ which is a clear contradiction . Thus only powers( $\neq 1$) of primes works.

We claim the solution is all prime powers which clearly work. Assume for the sake of contradiction that are two distinct primes that divide $n$. Call the smallest two prime divisors $p$ and $q$ with $p < q$. However, take the largest power of $p$ that is smaller than $q$. Let this be $p^k$, then take $\left(\frac nq, \frac n{p^k}, \frac{n}{p^{k-1}} \right)$ to get a contradiction.

The answer is all prime powers $n$, which clearly work. Otherwise, let $p$ and $q$ be the first and second smallest prime divisors of $n$ and let $v_p(n)=e$. Choosing a specific $i$ yields $$\frac{n}{q}\;|\;\frac{n}{p^{e+1}}+\frac{n}{p^{e}}\Rightarrow p^{e+1}\;|\;(p+1)q$$This can not hold, finishing the problem.

Answer : $n=p^x$ when $p$ is a prime and $x>1$ a natural number. It's so easy to see these form of solutions. We should look at for some other solutions. Firstly, assume that $n$ has two or more prime divisors. the smallest prime divisors of $n$ be $p<q$. Investigate the case that $p^x$ when $x>1$ and $p^x<q<p^{x+1}$. We can say, $1=d_1$, $p=d_2$, .... , $p^x = d_{x+1}$, $q =d_{x+2}$. From the given property, $d_x| (d_{x+1}+d_{x+2})$. $=$ $p^{x-1}|(p^x + q)$. Then, $(p^x+q)\equiv0\pmod{p}$ So, $p|q$ is impossible. From this, we can say that $x=1$. Investigate this case : If $x=1$, then $d_1=1 , d_2=p, d_3=q, .... , d_{k-2}=\frac{n}{q} , d_{k-1}=\frac{n}{p}$ and $d_k=n$. $d_{k-2}|(d_{k-1}+d_k) = \frac{n}{q}|(\frac{n}{p} + n)$. So, $p|(p+1)q$. Obviously, it's impossible. $n$ can't have two or more prime divisors. $\square$

The answer is $n=\boxed{p^k}$ for primes $p$ and positive integers $k\ge 2$ only. Clearly these work. Note that $d_{k-2}\mid d_{k-1}+d_k$, so $d_{k-2}\mid d_{k-1}$. Since $d_2d_{k-1}=d_3d_{k-2}=n$, we get that $d_2\mid d_3$. Let $p=d_2$, then let $d_3=xp$. If $x\ne p$, $x$ would have been a smaller divisor, so $x=p$ exactly. Then we may induct to show that all terms of $\{d_i\}$ besides $d_1$ are divisible by $p$, meaning that $n$ has no other prime divisors, as desired. $\blacksquare$

IMO 2023 p1 We claim that the answer is $n=p^k$, where $p$ is a prime. Let $p_1$ and $p_2$ be the first two primes of $n$. Now let $s$ be the largest number, such that $p_1^s < p_2$ So we have that \[\frac{n}{p_2},\frac{n}{p_1^s},\frac{n}{p_1^{s-1}}\]are consecutive divisors of $n$. Hence we must have that. $$\frac{n}{p_2}|(\frac{n}{p_1^s}+\frac{n}{p_1^{s-1}}) \implies p_1^s|p_2(p_1+1)$$Which is a contradiction, hence $n$ has less than 2 primes.

I claim the answer is only prime powers. It is easy to see that these work. To see nothing else works, assume for the sake of contradiction some $n$ with more than $1$ prime power worked. Then take the second smallest prime divisor, let it be $q$, and the largest divisor less than it, $p^{k}$, where $k$ can be $1$. We would then have $\frac nq \mid \frac{n}{p^{k - 1}} + \frac{n}{p^k}$, which is obviously not true by taking $\nu_p$ on both sides, the left side has $\nu_p(\frac nq) = \nu_p(n)$, right hand side has $\nu_p({\frac{n}{p^{k - 1}} + \frac{n}{p^k}}) = \nu_p({\frac{n}{p^k}})$, so the left side cannot divide the right.

Trivial NT on the IMO. Finally found some time to actually post my solution from the contest. We claim that the answer is all positive integers of the form $p^m$ for some prime $p$ and $m\ge 2$. These clearly all work since for all $1\le i \le m+1$ , $d_i=p^{i-1}$. Thus, \[d_i=p_{i-1}\mid p_i + p_{i+1}=d_{i+1}+d_{i+2}\]quite clearly. Now, we show that no other composite $n$ work. Say $n$ has atleast two distinct primes divisors, of which the smallest two are $p<q$. Note that we have $d_i = p^{i-1}$ for all $1\le i \le r$ for some $r \ge 2$ and $d_{r+1}=q$ as the divisors are arranged in increasing order. Now, if $n$ satisfies the desired condition we must have, \begin{align*} d_{m-r} &\mid d_{m-r+1} + d_{m-r+2}\\ \frac{n}{d_{r+1}} & \mid \frac{n}{d_r} + \frac{n}{d_{r-1}}\\ \frac{n}{q} & \mid \frac{n}{p^{r-1}} + \frac{n}{p^{r-2}}\\ p^{r-1} & \mid q + pq\\ p &\mid q+pq\\ p& \mid q \end{align*}which is a very clear contradiction since $q>p$ are both primes. Thus, there are no such $n$ of this form, which finishes the proof.

Clear answer is n =p^c where p is a prime Assume that n can have more than 1 prime factor and assume some prime q also divides n Well you get that p|q contradiction. Too easy for P1 tho

My first NT after decades..... IMO 2023 P1 wrote: Determine all composite integers $n>1$ that satisfy the following property: if $d_1$, $d_2$, $\ldots$, $d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \cdots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$. All numbers of the form $p^l$ where $p$ is a prime and $l>1$ work evidently. We now show these are the only ones. Let $m$ denote the number of divisors of $n$. Then we have $d_{i}d_{m+1-i}=n$. Thus, $$d_{i} \mid d_{i+1}+d_{i+2}$$$$\implies \frac{n}{d_{m-i+1}} \mid \frac{n}{d_{m-i}}+\frac{n}{d_{m-i-1}}$$$$\implies d_{m-i}d_{m-i-1} \mid d_{m-i+1}(d_{m-i}+d_{m-i-1})$$$$\implies \boxed{d_{i}d_{i+1} \mid d_{i+2}(d_{i}+d_{i+1})} [\because i \mapsto m-i-1]$$Now $i=1$ gives, $d_{1}d_{2} \mid d_{3}(d_{1}+d_{2}) \implies d_{2} \mid d_{3}$ since $d_{1}=1$. Also, $d_{2}$ is a prime number (say $p$). If $d_{3}$ has any prime factor $q$(say) other than $p$ then $d_{1}<q<d_{3};q \neq d_{2}$ which is a blatant contradiction. Thus $\boxed{d_{3}=p^{2}}$ as it can not be $p^e$ where $e \geq 3$ otherwise $d_{2} < p^{2} < d_{3}$ again, a contradiction. Now by an easy strong induction and similar argument we can conclude that $d_{i+1}=p^{i}$. Thus $n=p^{l}$ where $l>1$ and $p$ is a prime number. $\blacksquare$ ($\mathcal{QED}$)

Suppose $n$ has a prime divisors $\geq 2$. Say the least ones are $p$ and $q$ and $p$ be the minimum. Let the multiplicity of $p$ be $m$. At the $(k+1)$-th step, $k \leq m$: $p^{k-1} < q$, so $q$ can write $p^{k-1} \mid p^k + q \quad ( \rightarrow \leftarrow)$. Hence, $n = p^t$, $t \in \mathbb{N}$ and $p$ is prime.

Note that since $d_{k-2}\mid d_{k-1}+d_k$ and $d_{k-2}\mid n=d_k$, then $d_{k-2}\mid d_{k-1}$. Recall that $d_{k-2}=\frac{n}{d_3}$ and $d_{k-1}=\frac{n}{d_2}$, so $\frac{\frac{n}{d_2}}{\frac{n}{d_3}}\in\mathbb{N}\Longleftrightarrow\frac{d_3}{d_2}\in\mathbb{N}\Longleftrightarrow d_2\mid d_3$. But we also have $d_2\mid d_3+d_4$, so we must have $d_2\mid d_4\Longleftrightarrow\frac{\frac{n}{d_2}}{\frac{n}{d_4}}\in\mathbb{N}\Longleftrightarrow\frac{d_{k-1}}{d_{k-3}}\in\mathbb{N}\Longleftrightarrow d_{k-3}\mid d_{k-1}$. Again, we know that $d_{k-3}\mid d_{k-2}+d_{k-1}$ so we must have $d_{k-3}\mid d_{k-2}$. Repeating the process, eventually we'll end with $1=d_1\mid d_2\mid d_3\mid\dots\mid d_{k-1}\mid d_k=n$. Notice that for any natural number $n$, the three largest divisors of $n$ are $n,\frac{n}{p},\frac{n}{q}$ for prime $2\le p<q$ or $n,\frac{n}{p},\frac{n}{p^2}$ for prime $p\ge 2$. If the three largest divisors of $n$ are $n,\frac{n}{p},\frac{n}{q}$, then $\frac{\frac{n}{p}}{\frac{n}{q}}\in\mathbb{N}\Longleftrightarrow\frac{q}{p}\in\mathbb{Z}\Longleftrightarrow p\mid q$, contradiction. So the three largest divisors of $n$ are $n,\frac{n}{p},\frac{n}{p^2}$ for prime $p\ge 2$. In the similar spirit, we end up showing that $n=p_{k-1}$, which indeed works for any prime $p\ge 2$. Done. $\blacksquare$