Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

$d_k = {n\over d_1}$ $d_{k-1} = {n\over d_2}$ $d_1 = {n\over d_k}$ So we can write it like this: $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$ The Maximum that can be is: ${{n^2}\over 1*2}+{{n^2}\over 2*3}+\,\ldots\,+{{n^2}\over (n-1)n}$ We can see that the sum starting with the form: ${m\over {m+1}}{n^2}$ We check what happens if we add the next term: ${m\over {m+1}}{n^2} + {1\over {(m+1)(m+2)}}{n^2} = {{(m+1)^2}\over {(m+1)(m+2)}}{n^2} = {{m+1}\over {m+2}}{n^2}$ From here is easy to define that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$ $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$ is a divisor of ${n^2}$ when $n$ is a prime number: $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = n$ if $n$ isn't a prime we have at least 3 divisors of $n$ We proved that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$ So if it's a divisor of ${n^2}$ the maximum that can be: ${{n^2}\over d_2}$ $d_1 = 1$ $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = {{n^2}\over d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} > {{n^2}\over d_2}$ We see from here that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is a divisor of ${n^2}$ only if $n$ is a prime!

pavel25 wrote: $d_k = {n\over d_1}$ $d_{k-1} = {n\over d_2}$ $d_1 = {n\over d_k}$ So we can write it like this: $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$ The Maximum that can be is: ${{n^2}\over 1*2}+{{n^2}\over 2*3}+\,\ldots\,+{{n^2}\over (n-1)n}$ We can see that the sum starting with the form: ${m\over {m+1}}{n^2}$ We check what happens if we add the next term: ${m\over {m+1}}{n^2} + {1\over {(m+1)(m+2)}}{n^2} = {{(m+1)^2}\over {(m+1)(m+2)}}{n^2} = {{m+1}\over {m+2}}{n^2}$ From here is easy to define that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$ $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}$ is a divisor of ${n^2}$ when $n$ is a prime number: $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = n$ if $n$ isn't a prime we have at least 3 divisors of $n$ We proved that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k}<{n^2}$ So if it's a divisor of ${n^2}$ the maximum that can be: ${{n^2}\over d_2}$ $d_1 = 1$ $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$=${{n^2}\over d_1d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} = {{n^2}\over d_2}+{{n^2}\over d_2d_3}+\,\ldots\,+{{n^2}\over d_{k-1}d_k} > {{n^2}\over d_2}$ We see from here that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is a divisor of ${n^2}$ only if $n$ is a prime! I have asked this question as another link sorry for the same question <http://www.mathlinks.ro/Forum/viewtopic.php?p=531646#531646>

see i was right about the imo 2002 davron

Notice that because of the famous Cauchy Inequality applied to $d_1,...,d_{k-1}$ and $d_2,...,d_k$ we will have $d_1d_2+...+d_{k-1}d_k \le \left( d_1^2+...+d_{k-1}^2 \right)^2 \left( d_2^2+...+d_k^2 \right)^2$. Let $X=d_1^2+...+d_k^2$, we will have that $d_1d_2+...+d_{k-1}d_k \le \left( X-n^2 \right)^2 \left( X-1 \right)^2$ So, in order to finish the first part it is enough to prove $X^2-(n^2+1)X+(n^2-n^4) < 0$ which simplifies, by the general equation, to proving that $X < \displaystyle\frac{n^2+1+\sqrt{5n^4-4n^2+1}}{2}$ Since $\sqrt{5n^4-4n^2+1} > (2n^2-1)$, it is enough to prove that $X \le \displaystyle\frac{3n^2}{2}$ Let $n=p_1^{e_1}...p_m^{e_m}$, we will have $X=\prod_{i=1}^m (1+p_i^2+...+p_i^{2e_i})$ and so it will be enough to prove that, for each $i \le m$, if $p=p_i$ and $e=e_i$, $3p^{2e}/2 \ge 1+p^2+...+p^{2e}$ This is equivalent to $p^{2e}/2 \ge 1+p^2+...+p^{2e-2} = \displaystyle\frac{p^{2e}-1}{p^2-1}$ Notice that $p^2-1 \ge 4-1 = 3$ and so the above is true. So we are done with the first part. Now for the second part, assume that $X=d_1d_2+...+d_{k-1}d_k | n^2$. If $n$ is not prime then $k >2$ and so $X > d_kd_{k-1}=n^2/p$, where $p$ is the smallest prime divisor of $n$ (which exists because $n>1$). And so if $X | n^2$ then there exists a $q$ such that $qX=n^2$. Notice that $X < n^2$ by the first part and so $q > 1$. Since $q | n$, then $q \ge p$ and we get $pX \le n^2$ which is false because $X > n^2/p$. So $n$ must be prime, and it is easy to see $X=1*n|n^2$.

Part a: Notice that $d_i d_j = n $ if and only if $i+j = k+1$ So we have that $d_1d_2 +d_2d_3+...+d_{k-1}d_k < n^2 \iff$ $ d_{k-1}d_k (d_1d_2+...+d_{k-1}d_k) < n^2 (d_{k-1}d_k)$ Wich is equal to $n^2 + d_{k-1}d_k(d_2d_3+...+d_{k-1}d_k) < RHS$ Multiplying by the other terms we have that $ n^2(d_1d_2)(...)(d_{k-2}d_{k-1}) + n^2 (d_2d_3+...+d_{k-1}d_k) < n^2 (d_1d_2)(...)(d_{k-1}d_k)$ Wich is equal to $n^2(( \frac {(k-1) n^2} {d_{k-1}d_k})+(d_2d_3+...+d_{k-1}d_k)) < n^2 ((k-1) n^2))$ we have that clearly $d_id_{i+1}<n^2$ for every i So we just have to show that $\frac {(k-1) n^2} {d_{k-1}d_k} \le n^2 \iff $ $(k-1) \le d_{k-1}d_k$ We know that $ 1< d(n) < n$ so clearly this inequality holds ( $k-1<k=d(n)<n\le d_{k-1}n$) For part b: we have that the greatest divisor of $n^2$ is $\frac {n^2} {d_2} = d_{k-1} d_k $ And by part a we know that $n^2 > LHS \ge d_{k-1} d_k $ and equality holds only when n is a prime so thats the only posible case

Solution : Note that $\frac{d_1d_2 + d_2d_3 +\ldots + d_{k-1}d_k}{n^2} = \frac{1}{d_1d_2} + \frac{1}{d_2d_3} +\ldots + \frac{1}{d_{k-1}d_k}$. And use that $\frac{1}{d_1d_2} + \frac{1}{d_2d_3} +\ldots + \frac{1}{d_{k-1}d_k}$$\leq$ $ \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}+\ldots + \frac{1}{(k-1)k}$$\leq$ $ (1-\frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) +\ldots + (\frac{1}{k-1} - \frac{1}{k}) = 1 - \frac{1}{k} < 1$. So $\frac{d_1d_2 + d_2d_3 +\ldots + d_{k-1}d_k}{n^2}< 1$. $\Box$

My solution: We must to prove that $S=d_1d_2+...+d_{k-1}d_k=\frac{n^2}{d_1d_2}+...+\frac{n^2}{d_{k-1}d_k}<n^2.$ We know $n^2(\frac{1}{d_1d_2}+...\frac{1}{d_{k-1}d_k})\le n^2(\frac{1}{1\cdot 2}+...+\frac{1}{(k-1)\cdot k})=n^2(1-\frac{1}{k})<n^2.$ For part $b,$ we know $S\mid n^2,$ and $n=d_1d_k,n=d_2d_{k-1}\to n^2=d_2d_{k-1}d_k.$ Also we know $1<\frac{n^2}{S}\le \frac{n^2}{d_{k-1}d_k}=d_2.$ Equality: $S=d_{k-1}d_k\to k=2\to n=d_1\cdot d_2=d_2 $ where $d_2$ is prime.

(a) Clearly, $d_id_{k+1-i}=n$, so \[ d_id_{i+1} = \frac{n^2}{d_{k+1-i}d_{k-i}} = \frac{n^2}{d_{k+1-i}-d_{k-i}}\left(\frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}}\right) \le n^2\left(\frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}}\right), \]since $d_{k+1-i} - d_{k-i}\ge 1$. Summing, we get \[ \sum_{i=1}^n d_id_{i+1} \le n^2 \sum_{i=1}^n \frac{1}{d_{k-i}} - \frac{1}{d_{k+1-i}} = n^2\left(\frac{1}{d_1}-\frac{1}{d_k}\right)=n^2(1-1/n) < n^2.\](b) Let $S=d_1d_2+\cdots+d_{k-1}d_k$. First consider the case where $n$ is composite. We know $d_2$ is the smallest prime divisor of $n$. Let $d_2=p$. Since $d_{k-1}d_k=\tfrac{n}{p}\cdot n = \tfrac{n^2}{p}$, so $S>n^2/p$. But $p$ is also the smallest prime divisor of $n^2$, so $n^2/p$ is the largest proper divisor of $n^2$. Therefore, $S=n^2$, which is impossible, since it is at most $n^2(1-1/n)$. Hence, no composite $n$ works. Now consider the case where $n$ is prime. Then $d_1=1,d_2=n$, so $S=d_1d_2=n$, which divides $n^2$. Therefore, $S\mid n^2$ if and only if $n$ is prime.

We always have \begin{align*} d_k d_{k-1} + d_{k-1} d_{k-2} + \dots + d_2 d_1 &< n \cdot \frac n2 + \frac n2 \cdot \frac n3 + \dots \\ &= \left( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots \right) n^2 = n^2. \end{align*}This proves the first part. For the second, we claim that this only happens when $n$ is prime (in which case we get $d_1 d_2 = n$). Now assume $n$ is not prime (meaning $k \ge 2$) and let $p$ be the smallest prime dividing $n$. Then \[ d_k d_{k-1} + d_{k-1} d_{k-2} + \dots + d_2 d_1 > d_k d_{k-1} = \frac{n^2}{p} \]exceeds the largest proper divisor of $n^2$, but is less than $n^2$, so does not divide $n^2$.

orl wrote: Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$. Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$. Notice that \begin{align*} \sum^{k-1}_{i=1} d_id_{i+1} &=n^2 \cdot \left( \sum^{k-1}_{i=1} \frac{1}{d_id_{i+1}}\right) \\ & \le n^2 \cdot \left( \sum^{k-1}_{i=1} \frac{(d_{i+1}-d_i)}{d_id_{i+1}} \right) \\ &= n^2 \cdot \left(\sum^{k-1}_{i=1} \left(\frac{1}{d_i}-\frac{1}{d_{i+1}} \right) \right) \\ &= n^2 \cdot \left(1-\frac{1}{n}\right) <n^2, \end{align*}so the first assertion is true. Observe that $d_{k-1}=\frac{n}{p}$ where $p$ is the smallest prime factor of $n$; so if $k>2$ then $$1<\frac{n^2}{\sum^{k-1}_{i=1} d_id_{i+1}}<\frac{n^2}{d_{k-1}d_k}=p,$$contradicting that this quotient is a divisor of $n^2$. So, $k=2$ and $n=p$. For all prime numbers $n$, the assertion holds; hence these are the only solutions.

Note that \[S:=d_1d_2+\cdots+d_{k-1}d_k=n^2\left(\frac{1}{d_1d_2}+\cdots+\frac{1}{d_{k-1}d_k}\right)\le n^2\left(\left(\frac{1}{d_1}-\frac{1}{d_2}\right)+\cdots+\left(\frac{1}{d_{k-1}}-\frac{1}{d_k}\right)\right),\]so \[S\le n^2(1-1/n)=n^2-n<n^2,\]as desired. Now suppose that $S\mid n^2$. Any divisor of $n^2$ can be written as a product of two divisors of $n$, so suppose $S=d_ad_b$ where $b>a$. We'll show that this isn't possible for size reasons. To do so, we have the following ``smoothing'' lemma. Lemma: If $0<x\le z\le y$, then $xy<xz+zy$. Proof: Over the interval $[x,y]$, the extrema of the linear function $z(x+y)$ will be at the edges, so \[x^2+xy\le xz+zy\le y^2+xy,\]so certainly $xy<xz+zy$. $\blacksquare$ Suppose for now that $b-a\le 2$ Using the lemma, we see that \[d_ad_b<d_ad_{a+1}+d_{a+1}d_{a+2}+\cdots+d_{b-1}d_b,\]so $d_ad_b<S$. If $b-a=1$, then the only way $S=d_ad_{a+1}$ is if $n$ has only two factors, so $n$ is prime. Primes clearly work as $S=1\cdot p\mid p^2$. Thus, the answer is $\boxed{n\text{ is prime}}$.

Cute problem Notice that $d_k \leq \frac n1, d_{k-1} \leq \frac{n}2, d_{k-2} \leq \frac{n}3, $ and so on. Thus the sum $X = d_kd_{k-1}+d_{k-1}d_{k-2}+...+d_2d_1 \leq \frac{n^2}{1 \cdot 2} + \frac{n^2}{2 \cdot 3} +...+ \text{(k-1 terms)} < n^2(\frac 1{1 \cdot 2} +\frac 1{2 \cdot 3}+... \text{(infinite sum)})=n^2( \frac 11 -\frac 12 + \frac 12 - \frac 13 +\frac 13 -... )=n^2$. $=>$ $a$ part over. After a little playing around, or even otherwise, we see that the only $n$ that'll work should be prime $n$- once we realise this, it basically becomes a one-liner: consider $n$ non-prime with smallest divisor $p$. Notice that the largest divisor of $n^2<n^2$ must be $\frac{n^2}p$, and as our sum $X<n^2$ but $d_k d_{k-1} = \frac{n^2}p =>$ if there are more than $2$ factors, $X> \frac{n^2}p$ can't divide $n^2$. Verifying, we see that each $n$ prime indeed works, hence the only $n$ that work are primes.

First part $1=d_1<d_2<....<d_k=n$ are positive devisors of$n$ . Observe that$d_1d_k=n$ ,$d_2d_{k-1}=n$ and$d_id_{k+1-i}=n$. So,$d_1d_2+...+d_{k-1}d_k$. $=\frac{n}{d_k} \frac{n}{d_{k-1}} +\cdots +\frac{n}{d_1} \frac{n}{d_2}$. $= n^2(\frac{1}{d_1d_2} +\cdots +\frac{1}{d_kd_{k-1}})$.

. $\le n^2(\frac{d_2-d_1}{d_1d_2}+\cdots +\frac{d_k-d_{k-1}}{d_kd_{k-1}})$. $=n^2(\frac{1}{d_1}-\frac{1}{d_k})$. $<n^2$. As desired. Next part. $\boxed{\text{Claim}}$. Suppose,$S=d_1d_2+\cdots +d_kd_{k-1}$.if $S|n^2$ then $n$ must be prime . proof For contrary suppose $n$ is composite. and $n=pq$ where $\gcd (p,q)=1$. Let, $d_1=1,.....,d_{k-1}=p,d_k=pq$.where $q$ is a prime. Observe,that $d_kd_{k-1}<S<n^2$. Now ,$p^2q<S<p^2q^2$,since $S|n^2$ and less than$n^2$ thus$ S $ must be less or equal$p^2q$. Contradiction!!! This complete our proof.

By pairing up terms, we can see that \[\sum_{i=1}^{k-1}d_id_{i+1}=n^2\left(\sum_{i=1}^{k-1}\frac{1}{d_id_{i+1}}\right)\]A simple bound gives \[\sum_{i=1}^{k-1}\frac{1}{d_id_{i+1}}<\sum_{i=1}^{k-1}\frac{1}{i(i+1)}=1-\frac{1}{k}<1\]as desired. I claim that the sum divides $n^2$ only when $n$ is prime. It's easy to see that this is sufficient. To show it is necessary, suppose that $n$ is composite, or $k>2$. Let $p$ be the smallest prime divisor of $n$---clearly, $\frac{n^2}{p}$ is the largest proper divisor of $n^2$. If $k>2$, then \[\sum_{i=1}^{k-1}d_id_{i+1}<p+\frac{n^2}{p}+\text{blah}\]and since it is less than $n^2$ and greater than the largest proper divisor, it can not divide $n^2$. Thus, $n$ must be prime.

My first NT after some decades............. ISL 2002 N2 wrote: Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$. Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$. $$\sum_{i=1}^{k-1} d_id_{i+1}=n^2\left(\frac{1}{\sum_{i=1}^{k-1} d_id_{i+1}}\right)<n^2\left(\frac{d_{i+1}-d_i}{\sum_{i=1}^{k-1} d_id_{i+1}}\right)=n^2\left(1-\frac{1}{n}\right)<n^2$$Now when $n$ is a prime it is easy to check $\sum_{i=1}^{k-1} d_id_{i+1}|n^2$. Now suppose $n$ is not a prime and let $p$ be the smallest prime dividing $n$. So, $k\geq 3$. Then $\sum_{i=1}^{k-1} d_id_{i+1}=n^2\left(\frac{1}{\sum_{i=1}^{k-1} d_id_{i+1}}\right)<\frac{n^2}{\frac{n^2}{p}}=p$ contradicting the minimality of $p$. So, $k=1,2$ but $k\ne 1$ as $n\geq 2\implies k=2\implies n$ must be a prime. $\qquad\blacksquare$

I have extreme amounts of hating math - not because of this problem but because gradually math is more boring and i want to deviate to do CS or physics. $\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+...+\frac{n^2}{d_1d_2}\le n^2(\frac1{1\cdot 2}+\frac1{2\cdot 3}+...)=n^2(1-\frac1n)<n^2$ UNLESS n is prime, then the LHS=n, which works, and it's the only solution since if the smallest prime p dividing n is not n then to divide n we must have $LHS\le\frac{n^2}p$, contradiction since n and n/p already multiply to n^2/p in the sum, and we at least have the other factor 1 with the second smallest factor.

Notice that $$d_1d_2+d_2d_3+\cdots+d_{k-1}d_k = n^2\left(\frac 1{d_1d_2} + \frac 1{d_2d_3}+\cdots + \frac 1{d_{k-1}d_k}\right).$$On the other hand, as $\frac 1{d_1d_2} < \frac 1{d_1} - \frac 1{d_2}$, this is strictly less than $n^2$. On the other hand, as $d_{k-1}d_k = \frac{n^2}p$ where $p$ is the smallest prime divisor of $n$, it follows that the sum must equal exactly $\frac{n^2}p$. This occurs if and only if $n$ is prime.

Note that $d_k\le \frac n1$, $d_{k-1} \le \frac n2$, $d_{k-2} \le \frac n3$, $\ldots$, $d_2 \le \frac{n}{k-1}$, $d_1\le \frac{n}{k}$. In general, $d_i\le \frac{n}{k-i+1}$. Thus, we have the inequality$$d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k\le n^2\sum^{k-1}_{i=1}\frac 1{i(i+1)}.$$The summation telescopes into $1-\frac 1k$, so we have $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k\le n^2\left(1-\frac 1k\right)<n^2$, as desired. Now, I claim that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ divides $n$ only if $n$ is prime. Suppose the smallest divisor of $n$ other than $1$ is $p$. Obviously, $p$ is prime. Note also that $p$ is the smallest divisor of $n^2$ other than $1$. Then, we have that $d_2 = p$ and $d_{k-1} = \frac np$. We then have that our sum is equal to$$p+\frac{n^2}{p} + d_2d_3+d_3d_4\ldots + d_{k-2}d_{k-1}>\frac{n^2}{p}$$$$\implies \frac{n^2}{p+n^2+d_2d_3+d_3d_4+\ldots + d_{k-2}d_{k-1}}<p.$$However, if $p+\frac{n^2}{p} + d_2d_3 + d_3d_4+\ldots + d_{k-2}d_{k-1}\mid n^2$, then the LHS of the inequality is an integer, contradicting the minimality of $p$. Thus, the sum cannot divide $n^2$ if $n$ is composite, as desired. $\blacksquare$

Note that $d_id_{k-i+1} = n$ for all $1 \leq i \leq k$. Then, $D = d_1d_2 + d_2d_3 + \cdots \ d_{k-1}d_k$ has \[ D = n^2 (\frac{1}{d_{1}d_{2}} + \frac{1}{d_{2}d_{3}} + \cdots + \frac{1}{d_{k-1}d_{k}}) \]Since $1 = d_1 < d_2 < \cdots < d_k$, we have $i < d_i$ for all $1 \leq i \leq k$. This gives: \[ \frac{1}{d_{1}d_{2}} + \frac{1}{d_{2}d_{3}} + \cdots + \frac{1}{d_{k-1}d_{k}} \leq \sum_{i = 1}^{k-1}\frac{1}{(i)(i+1)} = \sum_{i = 1}^{k-1}\frac{1}{i} - \frac{1}{i+1}\]By telescoping, the above sum is equal to $1 - \frac{1}{k} < 1$, so $D < n^2 \cdot 1 = n^2$. Claim: $D \mid n^2$ if and only if $n$ is prime. It is easy to check that when $n$ is prime, $D = 1 \cdot n = n$, and trivially $n = D \mid n^2$. Otherwise, assume $n$ is not prime and proceed by contradiction. If $D \mid n^2$, then $Dp = n^2$ for some positive integer $p$. By the above, $p$ cannot be one, so it has to be $\geq$ the smallest non-one divisor of $n^2$. This is the smallest prime in the prime factorization of $n^2$, which is also the smallest prime in the prime factorization of $n$, and therefore $p \geq d_2$. This means that $d_2d_{k-1}d_k = n^2 \leq Dp$. Equality is only when $d_{k-1}d_k$ is the only term in $D$, i.e. $k = 2$, implying that $n$ is prime, which is a contradiction. Otherwise $n^2 < Dp$, contradicting $Dp \mid n^2$.

By considering the divisors in the form $\frac{n}{d}$ we get it suffices to prove $$\frac{1}{d_1d_2}+\frac{1}{d_2d_3}+\cdots +\frac{1}{d_{k-1}d_k}\leq 1$$Now note that $$\frac{1}{d_1d_2}+\cdots +\frac{1}{d_{k-1}d_k}\leq\frac{d_2-d_1}{d_1d_2}+\cdots \frac{d_k-d_{k-1}}{d_{k-1}d_k}=\frac{1}{d_1}-\frac{1}{d_2}+\cdots +\frac{1}{d_{k-1}}-\frac{1}{d_k}=1-\frac{1}{n}<1$$For the other part, note that $$n^2>d_1d_2+\cdots d_{k-1}d_{k}\geq d_{k-1}d_{k}=\frac{n}{d_2}\cdot n=\frac{n^2}{d_2}$$But $n^2$ and $\frac{n^2}{d_2}$ are the bigger and the second bigger divisors of $n^2$, and the equality is satisfied only when $n$ is prime

Solved with megarnie. We can rewrite the given sum as follows: \begin{align*} d_1d_2 + d_2d_3 + \cdots + d_{k-1}d_k &= \dfrac{n}{d_k}\cdot \dfrac{n}{d_{k-1}} + \dfrac{n}{d_{k-1}} \cdot \dfrac{n}{d_{k-2}} + \cdots + \dfrac{n}{d_2} \cdot \dfrac{n}{d_1} \\ &= \dfrac{n^2}{d_kd_{k-1}} + \dfrac{n^2}{d_{k-1}d_{k-2}} + \cdots + \dfrac{n^2}{d_2d_1} \\ &= n^2 \left(\dfrac{1}{d_kd_{k-1}} + \dfrac{1}{d_{k-1}d_{k-2}} + \cdots + \dfrac{1}{d_2d_1} \right) \end{align*}Thus, we just need to show that $$\dfrac{1}{d_kd_{k-1}} + \dfrac{1}{d_{k-1}d_{k-2}} + \cdots + \dfrac{1}{d_2d_1} < 1$$Indeed, we know that $d_2 \geq 2$, $d_3 \geq 3$, $\cdots$, so it follows that \begin{align*} \dfrac{1}{d_kd_{k-1}} + \dfrac{1}{d_{k-1}d_{k-2}} + \cdots + \dfrac{1}{d_2d_1} &\leq \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \cdots + \dfrac{1}{n(n-1)} \\ &= \dfrac{1}{1} - \dfrac 12 + \dfrac 12 - \dfrac 13 + \cdots - \dfrac{1}{n-1} + \dfrac{1}{n-1} - \dfrac 1n \\ &= 1 - \dfrac 1n \\ &< 1 \end{align*}and the desired inequality follows. Now, we claim that $d_1d_2 + d_2d_3 + \cdots + d_{k-1}d_k$ divides $n^2$ if and only if $n$ is a prime. We first show that all primes work. Indeed, the only divisors of $n$ are $1,n$, so the given sum is equal to $1 \cdot n = n \mid n^2$. To see that all composite $n$ do not work, let $p$ be the smallest prime that divides $n$. Then, $d_2 = p$, and $d_{k-1}d_k = \dfrac{n}{d_2} \cdot n = \dfrac{n^2}{p}$. For $d_1d_2 + d_2d_3 + \cdots + d_{k-1}d_k$ to divide $n$, we must have $$d_1d_2 + d_2d_3 + \cdots + d_{k-1}d_k = d_1d_2 + d_2d_3 + \cdots + \dfrac{n^2}{p} = \dfrac{n^2}{a}$$for some positive integer $a$. Since $n$ is composite and has at least $3$ positive divisors, we must have $$\dfrac{n^2}{p} < \dfrac{n^2}{a}$$which is equivalent to $p > a$. However, since $p$ is the smallest prime that divides $n$, $p$ is also the smallest prime that divides $n^2$, so such $a$ cannot exist, and we reach a contradiction if $n$ is composite. If $n$ is prime, then we could have $p=a$ as then we only have $2$ divisors, and the condition is satisfied as shown above. $\blacksquare$

We have $d_{k-1} \leq \frac{n}{2}$, $d_{k-2} \leq \frac{n}{3}$, $\dots$ so $$d_{1}d_{2} + d_{2}d_{3} + \dots + d_{k-1}d_{k} < n^2\left(1 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{3} + \dots\right) = n^2.$$For the second part, we claim the answer is when $n$ is prime. Let $p$ be the smallest prime factor of $n$. If $n = p$, then $d_1d_2 = n \mid n^2$, so prime $n$ work. If $n$ is not prime, and suppose $d_1d_2 + \dots d_{k-1}d_k = n^2/a$. Since $p$ is the smallest prime dividing $n$, we have $a \geq p$. But $$ d_1d_2 + d_2d_3 + \dots + d_{k-1}d_k > n^2\left(1\cdot \frac{1}{p}\right),$$so $a < p$, contradiction. Therefore, only prime $n$ work.

IMO 2002 p4 We first show that. $$d_{1}d_{2} + d_{2}d_{3} + \dots + d_{k-1}d_{k} < n^2.$$ So since we have that $d_i=\frac{n}{d_k-i}$, hence we have that. \begin{align*} d_1d_2 + d_2d_3 + \cdots + d_{k-1}d_k &= \dfrac{n}{d_k}\cdot \dfrac{n}{d_{k-1}} + \dfrac{n}{d_{k-1}} \cdot \dfrac{n}{d_{k-2}} + \cdots + \dfrac{n}{d_2} \cdot \dfrac{n}{d_1} \\ &= n^2 \left(\dfrac{1}{d_kd_{k-1}} + \dfrac{1}{d_{k-1}d_{k-2}} + \cdots + \dfrac{1}{d_2d_1} \right) \end{align*} So we need to show that. $$\dfrac{1}{d_kd_{k-1}} + \dfrac{1}{d_{k-1}d_{k-2}} + \cdots + \dfrac{1}{d_2d_1}<1$$. Which is easy. Now for the second part, we claim that it hold if and only if $n$ is a prime. Assume that $n$ is a prime, this case is trivial. Now assuming that $n$ is composite, then we have that. $$d_kd_{k-1}=\frac{n^2}{p}$$Where $p$ is the smallest prime of $n$. We want to show that $d_1d_2 + d_2d_3 + \dots + d_{k-1}d_k= \frac{n^2}{q}$ where $q$ is a divisor of $n$. But we have that $p<q$, hence only prime $n$ works.

Note that $d_k \leq n$, $d_{k - 1} \leq \frac{n}{2}$, $d_{k - 2} \leq \frac{n}{3}$, $\dots$, $d_{1} \leq \frac{n}{n}$. Observe that: $$d_1d_2 + d_2d_3 + ... + d_{k - 1}d_{k} \leq n^2(1 - \frac{1}{n})$$Now, since $1 - \frac{1}{n} < 1$, we are done. For the second part, note that if $n$ is composite, we have $d_{k - 1} = \frac{n}{p}$, where $p$ is the smallest prime divisor of $n$. Observe that $d_1d_2 + d_2d_3 + ... + d_{k - 1}d_{k} > \frac{n^2}{p}$, however this is the largest divisor of $n^2$, so we have a contradiction. If $n$ is prime, it works.

This expression is bounded by $n^2 (\frac 11 \frac 12 + \frac 12 \frac 13 + \cdots + \frac{1}{n - 1} \frac 1n) = n^2 - n $, as desired. We claim it is a divisor of $n^2$ only when $n$ is prime. It is easy to see all prime numbers work. To see that nothing else works, take the smallest prime divisor of $n^2$ as $p$, clearly the second largest divisor of $n^2$ is $\frac{n^2}{p}$, but $d_{k -1}d_k = \frac 1p n^2$, so we must have this being the only term in the summation, so $k = 2$ forces $n$ prime.

a) Notice that: $$d_1 d_2 + d_2 d_3 + \cdots + d_{k-1} d_{k} = n^2 \cdot \left( \frac{1}{d_1 d_2} + \frac{1}{d_2 d_3} + \cdots + \frac{1}{d_{k-1} d_{k}}\right).$$Notice that: $$\frac{1}{d_1 d_2} + \frac{1}{d_2 d_3} + \cdots + \frac{1}{d_{k-1} d_{k}} < \sum_{i=1}^{\infty} \frac{1}{i(i+1)} = 1$$ Thus, we have: $$d_1 d_2 + d_2 d_3 + \cdots + d_{k-1} d_{k} = n^2 \cdot \left( \frac{1}{d_1 d_2} + \frac{1}{d_2 d_3} + \cdots + \frac{1}{d_{k-1} d_{k}}\right) < n^2.$$ b) Note that: $d_1=1, d_2=p$ where $p$ is a the smallest prime factor of $n$. Then, notice that: $d_{k-1} d_k = \frac{n^2}{p}$ is the largest divisor of $n^2$. Thus, it should be the only term in the summation: $S = d_1 d_2 + d_2 d_3 + \cdots + d_{k-1} d_{k}$ inorder to have $S|n^2$. Therefore, $k=2$ which implies $n$ is a prime and we are done.

The first part follows from the very loose bound: \begin{align*} d_1 d_2 + d_2 d_3 + \dots + d_{k-1} d_k = n^2 \left( \frac{1}{d_1} \frac{1}{d_2} + \dots + \frac{1}{d_{k-1}} \frac{1}{d_k} \right) \\ < n^2 \left( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{(k-1) \cdot k} \right) \\ < n^2 \left( 1 - \frac{1}{k} \right) \\ < n^2 \\ \end{align*}The second part is only true when $n=\boxed{p}$ for some prime $p$. This works because $p | p^2$. If $n$ was compoite, then consider the smallest prime $q$ dividing it. The desired sum is then strictly greater than $n \cdot \frac{n}{q}$, implying the desired contradiction.

It amounts to proving $\sum \frac{1}{d_i.d_{i+1}}$ is less than $1$ where $i$ varies from $1$ to $k-1$. Note that $\frac{1}{d_id_{i+1}}$ is less than or equal to$\frac{1}{d_i} - \frac{1}{d_{i+1}}$ with equality holding iff $d_{i+1}-d_i=1$. Now obviously, for each $n$ there must be atleast one pair of consecutive divisors with a difference greater than $1$ so the above sum is less than $1- \frac{1}{n}$ which is less than $1$. Now notice that second largest divisor of $n^2$ is $\frac{n^2}{d_2}$ but this is $d_{k-1}d_k$ so the given sum is greater than this unless there is only one term in the sequence, that is, $k=2$. This implies $n$ must be a prime and all primes can easily be verified to work. So the answer to the second part is $n$ must be a prime.

We split the argument into two cases. Case 1 : If $n \ge 2 $ is a prime, $d_1=1$ and $d_2=n$ are all the divisors of $n$. Then, \[d_1d_2=n \nmid n^2\]so the inequality holds with $n$ a divisor of $n^2$. Case 2 : $n$ is composite. Then, $k >2$. Now note, \[d_1d_2+d_2d_3 + \dots + d_{k-1}d_k = \frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1}\]First note, \[\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1} > \frac{n^2}{d_2d_1} = \frac{n^2}{d_2}\]Further, since clearly $d_i \ge i$ for all $1\le i \le k$, \[\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1} \le \frac{n^2}{k(k-1)}+\frac{n^2}{(k-1)(k-2)}+\dots + \frac{n^2}{2 \cdot 1} = n^2 \left(1- \frac{1}{k}\right) < n^2\]Since $d_2$ is the smallest divisor of $n$, it must be prime and thus, it is also the smallest divisor of $n^2$ (every prime divisor of $n^2$ is a prime divisor of $n$). Thus, $\frac{n^2}{d_2}$ is the largest divisor of $n^2$ less than $n^2$. But since, \[\frac{n^2}{d_2}<\frac{n^2}{d_kd_{k-1}}+\frac{n^2}{d_{k-1}d_{k-2}}+\dots + \frac{n^2}{d_2d_1} < n^2\]it cannot be a divisor of $n^2$.