a) First take $n=6=2\times 3$ satisfies. If there are infinitely many lovely numbers , let the largest one be $n=d_1d_2\ldots d_k$. Let $d_{k+1}=n+1,N=n(n+1)=d_1d_2\ldots d_kd_{k+1}$. For $1\leq i\leq k: \frac{N+d_i}{d_i^2}=\frac{(d_1d_2\ldots d_k)^2}{d_i^2}+\frac{n+d_i}{d_i^2}$ is a integer. For $i=k+1: \frac{N+d_{k+1}}{d_{k+1}^2}=\frac{n(n+1)+n+1}{(n+1)^2}=1$. So $N$ is a lovely number larger than $n$. Contradiction.

b) First,we have $gcd(d_i,d_j)=1,\quad \forall 1\le i<j\le k$. This is because $d_i|\frac{n}{d_i}+1$,so $gcd(d_i,d_j)\le gcd(d_i,\frac{n}{d_i})=1$. In this way,we have $$d_i=a_i^2,\quad \forall 1\le i\le k$$We let $\overline{a_i^2}=\frac{n}{a_i^2}$. So $$a_i^2|\overline{a_i^2}+1,\quad \forall 1\le i\le k$$So we have \begin{align*} & \prod_{i=1}^k a_i^2|\prod_{i=1}^k \left (\overline{a_i^2}+1\right ) \\ \implies & \prod_{i=1}^k a_i^2|\left (\sum_{i=1}^k \overline{a_i^2}\right )+1 \end{align*}However,we have \begin{align*} & \frac{1}{a_1^2}+\frac{1}{a_2^2}+\cdots +\frac{1}{a_n^2} \\ < & \frac{1}{1\times 2}+\frac{1}{2\times 3}+\cdots +\frac{1}{n\times (n+1)} \\ < & 1 \end{align*}Which means $$ \prod_{i=1}^k a_i^2>\left (\sum_{i=1}^k \overline{a_i^2}\right )$$Contradiction!

If OTIS wants to make a unit ``Number Theory with Numbers and no Theory'' (as they already have ``Russian Combo'' for combi), then this problem is just perfect for the desired flavour! Congratulations to Novak, this is a brilliant problem. The brilliant solution to b) which I present here was done by a few Bulgarian contestants. The condition is equivalent to $d_i \mid \frac{n}{d_i} + 1$ for all $i$. a) Playing with small examples gives easily that $2$, $2 \cdot 3$ and $2\cdot 3 \cdot 7$ are lovely. For $k=4$ the relation $d_4 \mid d_1d_2d_3 + 1$ could now further support in taking $d_4 = d_1d_2d_3 + 1$ and continue in this manner. Formally, consider the sequence $a_1 = 2$, $a_{m+1} = a_1a_2\cdots a_m + 1$ and take $n = a_1a_2\cdots a_k$ for any $k$ (and $d_i = a_i$ for all $i$). Clearly $\frac{n}{a_1} $ is odd, so the condition holds for $i=1$. For $i\geq 2$ note that $a_1a_2 \cdots a_{i-1} \equiv -1 \pmod {a_i}$ and $a_{i+s} = a_1a_2 \cdots a_{i+s-1} + 1 \equiv 1 \pmod a_i$ for any $s>0$, so overall $\frac{n}{a_i} \equiv -1 \pmod {a_i}$, as desired. b) A starting observation is that gcd$(d_i, d_j) = 1$ for any $i$ and $j$ -- if not, then a common prime divisor $p$ would divide $\frac{n}{d_i}$ (which contains $d_j$ as a multiplier) and hence must divide $1$, contradiction. In particular, if their product $n$ is a square, we must have $d_i = x_i^2$ for some positive integers $x_i$. Now we have the relations $x_i^2 \mid \frac{n}{x_i^2} + 1$ and here comes the trick -- multiply them all! We obtain $n = (x_1x_2\cdots x_k)^2 \mid \prod_{i=1}^k\left(\frac{n}{x_i^2} + 1\right)$ and when we open all brackets at the right-hand side all terms apart from $1 + \sum_{i=1}^k \frac{n}{x_i^2}$ are divisible by $n=(x_1x_2\cdots x_k)^2$ (as each $\frac{n}{x_i^2}$ only misses $x_i^2$ but it appears at any $\frac{n}{x_j^2}$ for $j\neq i$). (Actually, I just figured out that there is no need to multiply them all -- one can just say that $\frac{n}{x_j^2}$ is divisible by $x_i^2$ for $j\neq i$ and from $x_i^2 \mid \frac{n}{x_i^2} + 1$ we obtain $x_i^2 \mid \sum_{i=1}^k \frac{n}{x_i^2} + 1$.) In particular, we must have $sn = 1 + \sum_{i=1}^{k} \frac{n}{x_i^2}$ for some positive integer $s$. One can nicely get rid of $s\geq 2$ by noting that since $n\geq 2^2 = 4$, we get $2 \leq \frac{1}{n} + \sum_{i=1}^k \frac{1}{x_i^2} < \frac{1}{4} + \frac{\pi^2}{6}$, contradiction. To deal with $s=1$ note further that we may assume that no $x_i$ is equal to $1$ (removing $x_i$-s equal to $1$ changes nothing) and $\sum_{t \geq 2} \frac{1}{t^2} = \frac{\pi^2}{6} - 1 < 1$ gives a contradiction! (If you wish to avoid using $\sum_{t=1}^{\infty} \frac{1}{t^2} = \frac{\pi^2}{6}$, then do some bounding like $\sum_{t=1}^{\infty} \frac{1}{t^2} < 1 + \sum_{t=2}^{\infty} \left(\frac{1}{t-1} - \frac{1}{t}\right) = 2$.)

a) It is easy to check that $n=6$ is a lovely number. Suppose that $n=d_1d_2\ldots d_k$ is a lovely number. We will prove that then the number $N=d_1d_2 \ldots d_k(d_1d_2\ldots d_k+1)$ is also a lovely number. Indeed observe that: $$ \frac{N}{d_i}+1 =\frac{n}{d_i} \cdot (d_1d_2 \ldots d_k+1) \equiv \frac{n}{d_i}+1 \equiv 0 \pmod {d_i} \implies d_i^2 \mid N+d_i$$for every $1 \le i \le k$. Moreover: $$ (d_1d_2\ldots d_k)+1 \mid \frac{N}{d_1d_2 \ldots d_k+1} +1= d_1d_2 \ldots d_k+1 $$This implies that $N$ is a lovely number too. Thus we can generate arbitrarily large lovely numbers as desired. b) Suppose there is a positive integer $n=d_1d_2 \ldots d_k$ which is a perfect square and is lovely too. Assume that then $d_i \neq 1$ for all $1 \le i \le k$ (because otherwise, we can neglect it, and consider number with $d_1$, $d_2, \ldots, d_{k-1}$. We proceed in several steps. Step 1: Numbers $d_1, d_2, \ldots, d_k$ are all perfect squares. Proof: We will prove that $\gcd(d_i, d_j) =1$, which will imply the conclusion. Suppose there exist prime $p$ dividing $\gcd(d_i, d_j)$. Consider: $$ p \mid d_i \mid \frac{n}{d_i}+1 = d_j \cdot \frac{n}{d_id_j}+1 $$As $p \mid d_j$ we see that we must have had $p \mid 1$, which is a contradiction. Step 2: $n \mid \frac{n}{d_1} + \frac{n}{d_2} + \ldots + \frac{n}{d_k}+1$. Proof: We know that $d_i \mid \frac{n}{d_i}+1$ since $n$ is a lovely number. Moreover for all $j \neq i$ we have $d_i \mid \frac{n}{d_j}$, so: $$ d_i \mid \frac{n}{d_1} + \frac{n}{d_2} + \ldots + \frac{n}{d_k}+1 $$Now we repeat this argument for every $1 \le i \le k$. Moreover, we have proven that $\gcd(d_i, d_j)=1$ for every $i \neq j$. Consequently: $$ n = d_1d_2 \ldots d_k \mid \frac{n}{d_1} + \frac{n}{d_2} + \ldots + \frac{n}{d_k}+1 $$ Step 3: Finish Proof: Suppose that $nT = \frac{n}{d_1} + \frac{n}{d_2} + \ldots + \frac{n}{d_k}+1$. This means that: $$ T = \frac{1}{d_1} + \frac{1}{d_2} + \ldots + \frac{1}{d_k} + \frac{1}{n} $$Remember that $d_1, d_2, \ldots d_k$ and $n$ are pairwise distinct perfect squares. This means that: $$ T = \frac{1}{d_1} + \frac{1}{d_2} + \ldots + \frac{1}{d_k} + \frac{1}{n}< \frac{1}{2^2}+ \frac{1}{3^2}+ \ldots = \frac{\pi^2}{6}-1 <1 $$This is a contradiction, so there are no integer $n$ which is both lovely and perfect square.

a) As the above posts mention, if $n=\prod\limits_{i=1}^{k} d_{i}$ is lovely, then $n=\prod\limits_{i=1}^{k} d_{i}\cdot \left(\prod\limits_{i=1}^{k} d_{i}+1\right)$ is lovely as well, so I won't focus on this easier part of the problem. b) Assume that such an $n$ exists. Then $\gcd(d_{i},d_{j})=1$ for $i\neq j$, so all $d_{i}$ must be pairwise different squares. Note that if $d_{i}=1$ we can just drop it (because we'll need to consider $d_{i}\geq 4$ in the bounds below) and $n$ would remain lovely. WLOG assume that $d_{1}<d_{2}<\dots <d_{k}$ and let \[c=\frac{\prod\limits_{i=1}^{k-1}d_{i}+1}{d_{k}}\in\mathbb{N}\]Then $d_{k}=\frac{\prod\limits_{i=1}^{k-1}d_{i}+1}{c}$. Substitute that in all the divisibilities $d_{i}\mid \frac{n}{d_{i}}+1$ to get: \[d_{i}\mid \frac{n}{d_{i}}+1= d_{k}\prod\limits_{\substack{ 1\leq j\leq k-1 \\ j\neq i}} d_{j}+1\Longrightarrow d_{i}\mid \frac{1}{c}\left(\prod\limits_{i=1}^{k-1}d_{i}+1\right)\prod\limits_{\substack{ 1\leq j\leq k-1 \\ j\neq i}} d_{j}+1\Longrightarrow d_{i}\mid \prod\limits_{\substack{ 1\leq j\leq k-1 \\ j\neq i}} d_{j}+c\]Since $d_{i}$ are pairwise coprime, notice that $c$ is uniquely determined $\pmod{\prod\limits_{i=1}^{k-1} d_{i}}$ due to CRT. However, notice that $C=n\left(1-\frac{1}{d_{1}}-\frac{1}{d_{2}}\dots -\frac{1}{d_{k-1}}\right)$ is an integer and satisfies all divisibilities as: \[\prod\limits_{\substack{ 1\leq j\leq k-1 \\ j\neq i}} d_{j}+C = \prod\limits_{\substack{ 1\leq j\leq k-1 \\ j\neq i}} d_{j}+n-\sum\limits_{t=1}^{k-1} \prod\limits_{\substack{ 1\leq s\leq k-1 \\ s\neq t}} d_{s}\equiv d_{i}\sum\limits_{\substack{1\leq t\leq k-1 \\ t\neq i}}\prod\limits_{\substack{1\leq s\leq k-1 \\ s\neq t, i}} d_{s}\equiv 0\pmod{d_{i}}\]Now we show that $\frac{n}{3}<C<n$. The second inequality is obviously true as $C=n\left(1-\frac{1}{d_{1}}-\frac{1}{d_{2}}\dots -\frac{1}{d_{k-1}}\right)<n\cdot 1=n$. For the first: \[C=n\left(1-\frac{1}{d_{1}}-\frac{1}{d_{2}}\dots -\frac{1}{d_{k-1}}\right)>n\left(1-\sum\limits_{i=2}^{\infty} \frac{1}{i^2}\right)=n\left(2-\frac{\pi^2}{6}\right)>\frac{n}{3}\]Hence we have the following two cases: Case 1. $c\equiv 1\pmod{\frac{n}{d_{k}}}$. Then because $c=\frac{\frac{n}{d_{k}}+1}{d_{k}}<\frac{n}{d_{k}}+1$, we must have $c=1$, but then \[n=\prod\limits_{i=1}^{k} d_{i} = (d_{k}-1)d_{k}\Longrightarrow (d_{k}-1)^2<n<d_{k}^2\text{, contradiction}\] Case 2. $c\not\equiv 1\pmod{\frac{n}{d_{k}}}$. Then $c=C$ as $c<\frac{n}{d_{k}}+1$, so $d_{k}=\frac{\frac{n}{d_{k}}+1}{c}<\frac{n+1}{\frac{n}{3}}<4$, which is impossible as $d_{i}\geq 4$ for all $i$.

certified $\zeta(2)$ moment Note that the condition is equivalent to $d_i \mid \frac{n}{d_i}+1$ for all $i$. For part (a), the answer is yes. We construct an infinite series $\{e_i\}_{i \geq 1}$ such that for any $k \geq 2$, $n(k):=\prod_{i=1}^k e_k$ is lovely. Take $e_1=2,e_2=3$, so $6$ is lovely as $2 \mid 3+1$ and $3 \mid 2+1$. Now for $k>2$, let $e_k=1+\prod_{i=1}^{k-1} e_i$, so for instance $e_3=7$ and $e_4=43$. Then clearly $e_k \mid \frac{n(k)}{e_k}+1$, and $e_i \mid \frac{n(k)}{e_i}+1 \iff e_i \mid \frac{n(k-1)}{e_i}+1$ for $i<k$, which is true by induction, so this indeed works. For part (b), the answer is no. First note that in general, we may freely add and remove factors of $1$ without changing the validity of a representation, so for ease of argument discard all factors of $1$. Note that if we have some $i\neq j$ with $\gcd(d_i,d_j)>1$, then $\gcd(d_i,d_j) \mid d_i$ and $\gcd(d_i,d_j) \mid \frac{n}{d_i}$, which means that $\gcd(d_i,d_j) \mid 1$: absurd. Thus we can say that if $n^2$ is lovely, then it can be decomposed into the coprime factors $c_1^2,\ldots,c_k^2$. The condition then gives us $$c_i^2 \mid \frac{n^2}{c_i^2}+1 \implies c_i^2 \mid \sum_{j=1}^k \frac{n^2}{c_j^2}+1,$$since clearly $c_i \mid \frac{n^2}{c_j^2}$ for $j \neq i$. Then, since all the $c_i^2$ are coprime, this implies that $$\prod_{j=1}^k c_j=n^2 \mid n^2\sum_{j=1}^k \frac{1}{c_j^2}+1 \implies n^2 \leq n^2\sum_{j=1}^k \frac{1}{c_j^2}+1 \implies \sum_{j=1}^k \frac{1}{c_j^2}+\frac{1}{n^2}\geq 1.$$However, because we removed all factors of $1$, $$\sum_{j=1}^k \frac{1}{c_j^2}+\frac{1}{n^2}\leq\sum_{i=2}^\infty \frac{1}{i^2}+\frac{1}{4}=\frac{\pi^2-6}{6}+\frac{1}{4}<1,$$which is a contradiction. $\blacksquare$

a) It is easy to check that all terms of the sequence $(n_k)_{k\geq 1}$ satisfying $n_1=6=2.3$ and $n_{k+1}=n_k(n_k+1)$, for all $k\geq 1$ is lovely by induction on $k$. b) Assume $n=\prod_{j=1}^k d_j$ is a lovely square. From $d_i^2\mid n+d_i$, we get $\nu_p(n)=\nu_p(d_i)$, so $\gcd (d_i,d_j)=1$ for all $i,j$. Moreover, $$\prod_{j=1}^k d_j^2 \mid \prod_{j=1}^k(n+d_j)\implies n\mid n\prod_{j=1}^k \frac1{d_j}+1.$$But $$\prod_{j=1}^k \frac1{d_j}\leq \sum \frac1{p_i^2}\leq \sum \frac1{(i+1)^2}<1-\frac1{n},$$absurd, so the assumption is false.