RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all divisors for some $m$ (except $1,m$). Find all such $n$. Do we have to find such $n$ in terms of a specific $m$ OR Is the problem as follows: RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all the divisors for some $m$ (except $1,m$). Find all such $n$.

My solution: Firstly,if n has an odd prime divisor p(p>=3). So p+1 divided by m so 2 is divided by m.(So a1+1=2(Contradition.) So n=2^k. Therefore:3;2^2+1;......;2^(k-1)+1 are all divisors of m. If m has 2 prime divisor p1;p2 so: p1*p2=(2^a+1)(2^b+1)=(2^c+1)(c>a;b) Or 2^(a+b)+2^a+2^b=2^c(Contradition) Therefore,n only has 1 prime divisor. This prime is =3. Therefore n=4.

tarzanjunior wrote: RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all divisors for some $m$ (except $1,m$). Find all such $n$. Do we have to find such $n$ in terms of a specific $m$ OR Is the problem as follows: RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all the divisors for some $m$ (except $1,m$). Find all such $n$. Second variant. anhtaitran wrote: p1*p2=(2^a+1)(2^b+1)=(2^c+1)(c>a;b) Why?

anhtaitran wrote: My solution: Firstly,if n has an odd prime divisor p(p>=3). So p+1 divided by m so 2 is divided by m.(So a1+1=2(Contradition.) So n=2^k. Let $n=2^{k+1}$ with $a_i=2^{i}$ for $1 \leq i \leq k$ If $k\geq 2$, $3$ and $5$ are factors of $m$ $\implies 15$ is also a factor of $m$, implying $14$ is a power of two, which is not true. Hence, $k=1$ since $n$ is composite. Hence, $n=2^2=4$

tarzanjunior wrote: $\implies 15$ is also a factor of $m$, implying $14$ is a power of two, which is not true. But if $m=15$?

RagvaloD wrote: tarzanjunior wrote: $\implies 15$ is also a factor of $m$, implying $14$ is a power of two, which is not true. But if $m=15$? Oops, I missed that case. So $ n$ can also be $8$

RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all the divisors for some $m$ (except $1,m$). Find all such $n$. Assume $k \geq 4$. Hence we get the chains $$n=a_1a_k=a_2a_{k-1}=\cdots$$$$m=(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1)=\cdots \implies u=a_1+a_k=a_2+a_{k-1}=\cdots$$Hence, the roots of the quadratic $x^2-ux+n=0$ are all the pairs $\{a_1,a_k\},\{a_2,a_{k-1}\}, \cdots$ which is a contradiction as $a_1<a_2<\cdots<a_k$. Hence, $k \le 3$. If $k=3$, then $n=a_2^2=a_1a_3$ and so $m=(a_2+1)^2=(a_1+1)(a_3+1) \geq (\sqrt{a_1a_3}+1)^2=(a_2+1)^2$, and thus equality holds and so $a_1=a_3$, a contradiction. Thus, $k=2$ or $1$. It is easy to see that the only possibility here are solutions having $4$ or $3$ distinct positive divisors. Hence, the only possible solutions are $n \in \ \{p^2, p^3, pq\}$ where $p \ne q$ are some primes. But since the aforementioned numbers are the divisors of $m$, hence we rule out some cases. $\bullet$ If $n=p^2$, then $m$ has only $3$ divisors, namely $1,p^2+1,m$ and so $p^2+1$ is a prime. Hence, $p$ must be even, and so $p=2$. Also, when $n=4$, we get $m=25$ which satisfies our conditons. $\bullet$ If $n=p^3$, then the divisors of $m$ are $1, p+1, p^2+1, m$, and so $p+1$ is a prime. Thus, $p=2$ giving $n=8$. In that case, $m=15$ works. $\bullet$ If $n=pq$, then the divisors of $m$ are $1, p+1, q+1, m$, and again we get $p=2$. Since $p \ne q$, $q>3$. But then $q+1$ has a prime divisor that is not $p+1=3$, a contradiction. Hence, the only possible solutions are $\boxed{n \in \{4,8\}}$

Nice problem! Clearly $a_1$ must be a prime, since it is the smallest divisor of $n$ apart from $1$, and $a_k = \tfrac{n}{a_1}$. Now $(a_1 + 1)(a_k + 1) = m \implies n + a_1 + a_k + 1 = m$. So $$n + p + \tfrac{n}{p} + 1 = m$$where $a_1 = p$ is a prime. Suppose that $n$ has another prime divisor $q$. Then we must have $a_r = q$ and $a_{k-r+1} = \tfrac{n}{q}$. So $$(a_r + 1)(a_{k-r+1} + 1) = m \implies n + q + \tfrac{n}{q} + 1 = m \implies p + \tfrac{n}{p} = q + \tfrac{n}{q}$$Simplifying, we get $n = pq$. Hence, $k = 2, a_1 = p < a_2 = q$. But now $a_1 + 1$ is also prime as it is the smallest divisor of $m$, so $p=2$. $q + 1$ cannot be prime, as $q$ is a prime and $q \neq 2$. So the only other option is $q + 1 = (p+1)^2 = 9 \implies q = 8$, absurd. So $n$ cannot have more than one prime divisor, but also this prime divisor must be equal to $2$, as already established. $n = 2^t$. Suppose $t \geqslant 5$. Then the divisors for $m$ must be $3<5<9<17<33<\cdots 2^{t-1} + 1$ which implies that $33$ is a divisor of $m$ but $11$ is not a divisor of $m$, impossible. Hence $t < 5$. Checking all values of $t$ gives $t \in \{ 2, 3 \} \implies \boxed{n \in \{4, 8 \} }$.

Claim : $n$ is even and $m$ is odd If $n$ is odd then $m$ is even but it does not have $2$ as a divisor. This is a contradiction. Hence $n$ is even and $m$ is odd. We inspect the cases where $k\leq 2$ $k=1$: Since $n$ and $m$ both have $3$ divisors each and they are perfect squares we conclude that both $n$ and $m$ are squares of primes. Hence $a_1$ and $a_1 + 1$ are primes, implying that $a_1 = 2$. Hence $n=4$. $k=2$ : We know that $a_1 = 2$. If $n$ has an odd divisor, then $m$ has an even divisor, which is contrary to the above claim. Hence $n$ does not have an odd divisor. Since $n$ is a perfect cube $n=8$. $k \geq 2$ If $k$ is even we know that \[a_1a_k = a_2a_{k-1} = \cdots = a_{k/2}a_{k/2 + 1}\]\[(a_1+1)(a_k+1) = (a_2 + 1)(a_{k-1} + 1) = \cdots = (a_{k/2})(a_{k/2 + 1} + 1)\]Combining the above two yeilds \[a_1 + a_k = a_2 + a_{k-1} = \cdots = a_{k/2} + a_{k/2 + 1}\] So $a_1 = 2, a_2 = 2+e_1, \cdots a_k = 2 + e_{k-1}$ where $e_i$ are nonzero because the divisors are increasing. \begin{align*} &4 + e_{k-1} = 4 + e_1 + e_{k-2} \\ &\Longrightarrow e_{k-1} = e_1 + e_{k-2} \\ &4 + 2e_{k-1} = 4 + 2e_{k-2} + 2e_1 + e_1e_{k-2} \\ &\Longrightarrow e_1e_{k-2} = 0 \end{align*}This is a contradiction. If $k$ is odd we know that both $n$ and $m$ are perfect squares. Let $\sqrt{n} = a_{\ell}$. We know that \[2 + a_k = 2a_{\ell}\]\[2a_k = a_{\ell}^2\]$2a_k = 4a_{\ell} - 4 = a_{\ell}^2 \Longrightarrow a_{\ell} = 2$ which is a contradiction since the divisors are ascending. We conclude that $k\leq 2$. Hence the only solutions are $\boxed{n=4}$ and $\boxed{n=8}$

Solution. We have $n=4,8$ as our only solutions. For now assume $k \ge 4$ then notice that \begin{align*} m &=(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1) \\ &=(a_1+1)(\tfrac{n}{a_1}+1)=(a_2+1)(\tfrac{n}{a_2}+1) \\ &=\cancel{n+1}+a_1+\tfrac{n}{a_1}=\cancel{n+1}+a_2+\tfrac{n}{a_2} \\ \end{align*}Move everything and solve to obtain $a_1a_2=n=a_1a_{k-1}$ which gives $a_1=a_{k-1}$ which is a contradiction. $\square$ We conclude $n \le 3$. We divide this into the following cases. If $k=3$ we have $\{1,a_1,a_2,a_3,n\}$ as divisors of $n$ which counts $5$. Hence $n=p^4$ for some $p$ but then notice that $\{1,p+1,p^2+1,p^3+1,m\}$ forms the list of divisors of $m$. Hence we obtain $(p+1)(p^3+1)=m=q^4$ which gives $p+1=q$ by which we get $p=2$ but then $9 \neq q^3$. Hence no solutions. If $k=2$ we have $\{1,a_1,a_2,n\}$ as divisors of $n$ which counts $4$. Hence $n=p^3$ or $n=pq$. In former case we have $\{1,p+1,p^2+1,m\}$ forms the list of divisors of $m,$ by which we get $(p+1)(p^2+1)=m=rs$ or $r^3$ in latter we have contradiction similar to one previously mentioned, otherwise we have $p=2$ and $s=5$ with $r=3$ as our solutions for above equation. Consequently, we get $n=8$ as our desired solution, where $m=3 \cdot 5=15$. In latter case we get $(p+1)(q+1)=m=rs$ or $r^3,$ if $m=rs$ we get $p=q=2,$ not possible, otherwise we have $p=2$ and $r=3$ from $p+1=r$ but then $q+1=r^2$ which gives $r=2$ and $q=3$. Contradiction. So only one solution as $n=8$ If $k=1$ we have $\{1,a_1,n\}$ as divisors of $n$ which counts $3$. Hence $n=p^2$ then $\{1,p+1,m\}$ forms divisor set of $m,$ we get, therefore, $p+1=q$ which gives $q=3$ and $p=2$ giving $n=4$ and $m=9$ We get $n=8$ and $n=4$ as our only final solutions. $\blacksquare$

I claim the only solutions are $n=4,8$, which are easy to check. Now, assume $k \geq 4$. Notice that \[n=a_1a_k=a_2a_{k-1}=a_3a_{k-2}=\dots\]and \[m=(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1)=(a_3+1)(a_{k-2}+1)=\dots. \]Expanding the products, \[a_1a_k+a_1+a_k+1=a_2a_{k-1}+a_2+a_{k-1}+1=a_3a_{k-2}+a_3+a_{k-2}+1=\dots.\]However, all the $a_ia_{k+1-i}$ are equal, so we have \[a_1+a_k=a_2+a_{k-1}=a_3+a_{k-2}=\dots.\]Looking at each two consecutive terms in the equality chain, \[a_k-a_{k-1}=a_2-a_1\]\[a_{k-1}-a_{k-2}=a_3-a_2\]\[\vdots\]Looking at $a_{k-1}-a_{k-2}=a_3-a_2$, this means that the difference between $a_3$ and $a_2$ is the same as the difference between $a_{k-1}$ and $a_{k-2}$. Let $x$ be the difference, so we have $a_3=a_2+x$ and $a_{k-2}=a_{k-1}-x$. Multiplying them, $a_3a_{k-2}=a_2a_{k-1}-a_2x+a_{k-1}x-x^2.$ Since $a_3a_{k-2}=a_2a_{k-1}$ and $x \neq 0$, \[a_2+x=a_{k-1}.\]But recall that $a_2+x=a_3$, so $a_{k-1}=a_3$. This means that $k=4$. Since $k=4$, the divisors of $n$ are $1, a_1, a_2, a_3, a_4, n$ and the divisors of $m$ are $1, a_1+1, a_2+1, a_3+1, a_4+1, m$. Since $a_1$ and $a_1+1$ are both prime, $a_1=2$. This means that $n$ is even and $m$ is odd. Since $a_1+a_4=a_2+a_3$ and $(a_1+1)+(a_4+1)=(a_2+1)+(a_3+1)$, $\frac{n}{2}+2=a_2+a_3$ and $\frac{m}{3}+3=a_2+a_3+2$. Combining them, $\frac{m}{3}=\frac{n}{2}+1$. This gives $2m=3n+1$, but the left side is even while the right side is odd, so there $k=4$ is not possible. Hence, we've concluded that $k$ cannot be 4 or greater. If $k=3$, then $n$ has 5 factors and 1 prime factor. The divisors of n are $1, p, p^2, p^3, n$ and the divisors of $m$ are $1, p+1, p^2+1, p^3+1, m$. We need $(p+1)(p^3+1)=(p^2+1)^2 \implies p^4+p^3+p+1=p^4+2p^2+1 \implies p^3-2p^2+p=0$. But this gives $p=0,1$ which cannot happen. Hence $k \neq 3$. If $k=2$, then the factors of $n$ are $1, a_1, a_2, n$ and the factors of $m$ are $1, a_1+1, a_2+1, m$. Both $a_1$ and $a_1+1$ must be prime, so $a_1=2$. Then, $a_2$ and $a_2+1$ can both be prime or one is a square of $a_1$ or $a_1+1$ respectively. However, the only possibility is if $a_2=a_1^2=4$. This gives $a_1=2, a_2=4$, so $n=8$. If $k=1$, then $a_1$ and $a_1+1$ must both be prime. This gives $a_1=2$ and $n=4$. We've exhausted all cases, so $n=4,8.$

Wizard_32 wrote: RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all the divisors for some $m$ (except $1,m$). Find all such $n$. Assume $k \geq 4$. Hence we get the chains $$n=a_1a_k=a_2a_{k-1}=\cdots$$$$m=(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1)=\cdots \implies u=a_1+a_k=a_2+a_{k-1}=\cdots$$Hence, the roots of the quadratic $x^2-ux+n=0$ are all the pairs $\{a_1,a_k\},\{a_2,a_{k-1}\}, \cdots$ which is a contradiction as $a_1<a_2<\cdots<a_k$. Hence, $k \le 3$. If $k=3$, then $n=a_2^2=a_1a_3$ and so $m=(a_2+1)^2=(a_1+1)(a_3+1) \geq (\sqrt{a_1a_3}+1)^2=(a_2+1)^2$, and thus equality holds and so $a_1=a_3$, a contradiction. Thus, $k=2$ or $1$. It is easy to see that the only possibility here are solutions having $4$ or $3$ distinct positive divisors. Hence, the only possible solutions are $n \in \ \{p^2, p^3, pq\}$ where $p \ne q$ are some primes. But since the aforementioned numbers are the divisors of $m$, hence we rule out some cases. $\bullet$ If $n=p^2$, then $m$ has only $3$ divisors, namely $1,p^2+1,m$ and so $p^2+1$ is a prime. Hence, $p$ must be even, and so $p=2$. Also, when $n=4$, we get $m=25$ which satisfies our conditons. $\bullet$ If $n=p^3$, then the divisors of $m$ are $1, p+1, p^2+1, m$, and so $p+1$ is a prime. Thus, $p=2$ giving $n=8$. In that case, $m=15$ works. $\bullet$ If $n=pq$, then the divisors of $m$ are $1, p+1, q+1, m$, and again we get $p=2$. Since $p \ne q$, $q>3$. But then $q+1$ has a prime divisor that is not $p+1=3$, a contradiction. Hence, the only possible solutions are $\boxed{n \in \{4,8\}}$ GoodOne

The answers are $n = 4, 8$. I claim that $n$ must be in the form $2^x$ for nonnegative integer $x$. Suppose not. Then there exists an odd prime $p$ such that $p|n$. Then, we get $p+1|m$. Since the list $a_1+1, a_2+1, ..., a_k+1$ contains all divisors of $m$ other than $1$ and $m$, $2$ must be included in the list, and one of $a_1, a_2, ..., a_k$ must be $1$. But this is impossible due to the set conditions. Thus $n$ is in the form $2^x$. Now, we get $a_1 = 2$, $a_2 = 4$, ..., $a_k = 2^{x-1}$, so all members of the list $$2+1, 2^2+1, 2^3+1, ..., 2^{x-1}+1$$are divisors of $m$. We check that $x \geq 4$ doesn't work because if so, our list of divisors of $m$ contains $3, 5, 9, ....$. But then $15$ must be included in the list, but clearly it is not able to be expressed as $2^y + 1$. Therefore $x<4$. Testing $x = 0, 1, 2, 3$ yields that $x = 2, 3$ work, which yield $n = 4, 8$. $\blacksquare$ I am curious: what if the conditions were set as $1 = a_1 < a_2 < \cdots < a_n = n$...

I feel like the (except 1,$m$) condition is weird :thonk:

superagh wrote: I am curious: what if the conditions were set as $1 = a_1 < a_2 < \cdots < a_n = n$... If $p=a_2$ then we must have that $2$ and $p+1$ must be the first nontrivial divisors of $m$, and so $p+1$ must be either a prime or $4$, but then we must have $p=2$ or $p=3$ In the first case we must have $n+1=\frac{m}{2}$ and $\frac{n}{2}+1=\frac{m}{3}$ but this would lead to $n=-\frac 12$ which is a contradiction. In the second case we would have $n+1=\frac{m}{2}$ and $\frac{n}{2}+1=\frac{M}{4}$ but this leads to a contradiction since $n+1=\frac{m}{2}=2\frac{m}{4}=2(\frac{n}{2}+1)=n+2$

RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all divisors for some $m$ (except $1,m$). Find all such $n$. if we change question such 1=a1<.....ak=n what differences will be happening?

Note that $m$ has all factors other than $1$ greater than $2$. Hence $m$ is odd, so every divisor of $n$ except $1$ is even and $n$ is a power of $2$. If $n > 8$, $n$ has at least $5$ factors so $n\neq 15$. But $2+1 = 3 | n$ and $4+1 = 5 | n$ so $15 | n$, absurd since $15$ is not one more than a power than $2$. It follows that $n\in\{4,8\}$, which both work.

mst.4921 wrote: RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all divisors for some $m$ (except $1,m$). Find all such $n$. if we change question such 1=a1<.....ak=n what differences will be happening? Let $p$ be the least prime factor of $n$, and let $n = dp$. Since $1 + 1 = 2 | m$ and $dp+1$ is the greatest factor of $m$ other than $m$ itself, we must have $m = 2(dp+1)$. It follows that $$d + 1 | 2dp + 2\implies d + 1 | 2(p-1)$$ Since $p$ is the least prime factor of $n$, we must have either $d\ge p$ or $d = 1$. Case $1$: $d = 1$. Then $n = p$ and $m = 2(p+1)$ with no factors other than $1$, $2$, $p+1$, $2(p+1)$. It follows that either $p+1$ is prime or $p+1 = 4$, so $p = 2,3$ which both work. Case $2$: $d\ge p$. Since $d + 1 | 2(p-1)$, we must have $d = 2p - 3$. It follows that $n = 2p^2 - 3p$ and $m = 2(2p^2 - 3p + 1) = (2p-2)(2p-1)$. Now, $d+1 = 2p-2$ is the greatest factor of $m$ less than $n+1$, implying $2p - 1 = n+1 = (2p-1)(p-1)$, so $p=2$ and $n=2$. In summary, the only solutions are $n = 1,2,3$.

mst.4921 wrote: RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all divisors for some $m$ (except $1,m$). Find all such $n$. if we change question such 1=a1<.....ak=n what differences will be happening? Assume $k \geq 4$ Notice that we have \[ n=a_1a_k = a_2a_{k-1} = ... \] Similarly, we have \begin{align*} m &= (a_1+1)(a_k+1) \\ &= (a_2+1)(a_{k-1}+1) ... \end{align*} Therefore, setting any two of them equal, we have \begin{align*} (a_1+1)(a_k+1) &= (a_2+1)(a_{k-1}+1) \\ a_1a_k+a_1+a_k+1 &= a_2a_{k-1}+a_2+a_{k-1}+1 \\ n+a_1+a_k+1 &= n+a_2+a_{k-1}+1 \\ a_1+a_k &= a_2+a_{k-1} \\ 1+n &= a_2+a_{k-1} \\ 1+a_2a_{k-1} &= a_2+a_{k-1} \\ (a_2-1)(a_{k-1}-1) = 0 \end{align*} Thus, either $a_2 = 1$ or $a_{k-1}=1$, contradiction. Therefore, $k=\{2,3 \}$ Trivially, $n=1$ works. Case 1: k=2 This means $n$ is prime or 1, so call it $p$, and we have $2, p+1$ the set of all proper divisors for some $m$. We must have $p+1$ even, so $p+1 = 4$ and $p=3, $ meaning that $n=3.$ Also, $n=2$ works. Case 2: k=3 We must have $n=p^2$ and so $\{2, p+1, p^2+1 \}$ form the entire set of proper divisors of $m$. Since $p+1$ is even except for $p=2,$ which is impossible, or $p+1=4$, which again is a contradiction (since the set of proper divisors of m needs to be greater). Therefore, we have no cases here. In conclusion, $n=\boxed{1, 2, 3}$

$$n=a_{i}a_{k+1-i}$$

Probably the 7th time but finally done with 0 hints. Also mobile writeups ARO 2017 Day 1 Grade 10 #1 wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all divisors for some $m$ (except $1,m$). Find all such $n$. Claim :$k>4$ isnt possible. Proof :Just note that the quadratic $x^2-(m-n+1)x+n$ can't have $>2$ roots,done $\blacksquare$ Claim :$k \ne 4,3$ Proof :For $k=4$,we get $a_1,a_2,a_3,a_4$ as the distinct divisors of $n$ excluding $\{1,n\}$.Now note that, $$(a_1+1)(a_4+1)=m=(a_2+1)(a_3+1) \implies a_1+a_4=a_2+a_3=m+n-1 \implies a_2-a_1=a_4-a_3=\frac{n}{a_1}-\frac{n}{a_2} \implies n=a_1a_2$$Thus we have a contradiction! For $k=3$,if $a_1,a_2,a_3$ become the distinct divisors of $n$ excluding $\{1,n\}$.Thus we have$$a_2^2=a_1a_3,(a_2+1)^2=(a_1+1)(a_3+1) \implies 2a_2=a_1+a_3$$Or $\{a_2,a_2,a_3\}$ forms both A.P and G.P.Since they are distinct its not possible and thus,contradiction $\blacksquare$ For $k=2,n=pq,p^3$ where $p,q$ are primes .Note that for $n=pq$,one of $\{p,q\}$ must be odd $\implies m$ has an extra power of $2$ leading to an extra divisor other than $p+1,q+1$.This is not possible $\implies n=p^3$,with $p$ prime.So this gives us $2$ conditions $(p,p+1)$ both are primes. $(p^2+1,p+1)$ are co-prime We use the first condition to get $p=2 \implies (n,m)=(8,15)$ For $k=1$ we get $n=p^2$,for some prime $p$.So this gives $m=(p+1)^2$ for a prime $p+1$.This again gives $p=2 \implies (n,m)=(4,9)$.The End $\blacksquare$

green_leaf wrote: Nice problem! Clearly $a_1$ must be a prime, since it is the smallest divisor of $n$ apart from $1$, and $a_k = \tfrac{n}{a_1}$. Now $(a_1 + 1)(a_k + 1) = m \implies n + a_1 + a_k + 1 = m$. So $$n + p + \tfrac{n}{p} + 1 = m$$where $a_1 = p$ is a prime. Suppose that $n$ has another prime divisor $q$. Then we must have $a_r = q$ and $a_{k-r+1} = \tfrac{n}{q}$. So $$(a_r + 1)(a_{k-r+1} + 1) = m \implies n + q + \tfrac{n}{q} + 1 = m \implies p + \tfrac{n}{p} = q + \tfrac{n}{q}$$Simplifying, we get $n = pq$. Hence, $k = 2, a_1 = p < a_2 = q$. But now $a_1 + 1$ is also prime as it is the smallest divisor of $m$, so $p=2$. $q + 1$ cannot be prime, as $q$ is a prime and $q \neq 2$. So the only other option is $q + 1 = (p+1)^2 = 9 \implies q = 8$, absurd. So $n$ cannot have more than one prime divisor, but also this prime divisor must be equal to $2$, as already established. $n = 2^t$. Suppose $t \geqslant 5$. Then the divisors for $m$ must be $3<5<9<17<33<\cdots 2^{t-1} + 1$ which implies that $33$ is a divisor of $m$ but $11$ is not a divisor of $m$, impossible. Hence $t < 5$. Checking all values of $t$ gives $t \in \{ 2, 3 \} \implies \boxed{n \in \{4, 8 \} }$. amazing solution but a quick edit $k \le 2$

Claim: Odd prime p does not divide n Proof: Let's assume, for the sake of contradiction, that p|n. Then, p+1|m, so 2|m, which is a contradiction, because $1<a_1$. Therefore, $n=2^k$. Now, note that 4 works and 8 works. Assume $k >3$ now. Then, we get that $2+1,4+1,..., 2^{k-2}+1, 2^{k-1}+1$ divides m. Thus, $3(2^{k-1}+1) = 5(2^{k-2} +1)$, and this equation simplifies to $3(2^{k-2}) = 5(2^{k-3})+1$ which has no solutions except when k=3, but since $k>3$, this is a contradiction. This implies $n=4,8$.

Wizard_32 wrote: RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all the divisors for some $m$ (except $1,m$). Find all such $n$. Assume $k \geq 4$. Hence we get the chains $$n=a_1a_k=a_2a_{k-1}=\cdots$$$$m=(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1)=\cdots \implies u=a_1+a_k=a_2+a_{k-1}=\cdots$$Hence, the roots of the quadratic $x^2-ux+n=0$ are all the pairs $\{a_1,a_k\},\{a_2,a_{k-1}\}, \cdots$ which is a contradiction as $a_1<a_2<\cdots<a_k$. Hence, $k \le 3$. If $k=3$, then $n=a_2^2=a_1a_3$ and so $m=(a_2+1)^2=(a_1+1)(a_3+1) \geq (\sqrt{a_1a_3}+1)^2=(a_2+1)^2$, and thus equality holds and so $a_1=a_3$, a contradiction. Thus, $k=2$ or $1$. It is easy to see that the only possibility here are solutions having $4$ or $3$ distinct positive divisors. Hence, the only possible solutions are $n \in \ \{p^2, p^3, pq\}$ where $p \ne q$ are some primes. But since the aforementioned numbers are the divisors of $m$, hence we rule out some cases. $\bullet$ If $n=p^2$, then $m$ has only $3$ divisors, namely $1,p^2+1,m$ and so $p^2+1$ is a prime. Hence, $p$ must be even, and so $p=2$. Also, when $n=4$, we get $m=25$ which satisfies our conditons. $\bullet$ If $n=p^3$, then the divisors of $m$ are $1, p+1, p^2+1, m$, and so $p+1$ is a prime. Thus, $p=2$ giving $n=8$. In that case, $m=15$ works. $\bullet$ If $n=pq$, then the divisors of $m$ are $1, p+1, q+1, m$, and again we get $p=2$. Since $p \ne q$, $q>3$. But then $q+1$ has a prime divisor that is not $p+1=3$, a contradiction. Hence, the only possible solutions are $\boxed{n \in \{4,8\}}$ You wrote $\bullet$ If $n=p^2$, then $m$ has only $3$ divisors, namely $1,p^2+1,m$ and so $p^2+1$ is a prime} but if $n=p^2 \implies n=1 \times p^2 = p\times p$ So all the divisors of $m$ would be $1,(1+1),p+1,p^2+1,m$ But you wrote it $3$ Confused

Note that n = 2^a for some natural a. This is true since it is impossible to include all of m's factors if m is even, since all a_i are > 1, meaning we will always miss 2 as a factor of m in the sequence {a_1+1, a_2+1, a_3+1, ... , a_k+1}. So, all a_i must be even in order for all the factors of m to be odd, producing an odd m overall. All a_i being even is the same as being a power of 2. Then we can consider the powers of 2 as values of n. n = 4, 8 works. Beyond that, 15 will always be missed as a factor of m for m = 16, 32, 64, etc. Thus the only solutions are 4 and 8.

Solved with blueberryfaygo_55. The answer is $\boxed{4,8}$. For $n = 4$, take $m = 9$ and for $n = 8$, take $m = 15$. Now we prove they are the only solutions. We see that $a_1$ and $a_1 + 1$ are both primes, as they are the smallest divisors of $m,n$, respectively. Therefore, $a_1 = 2$. This implies that $2\nmid m$, so all divisors of $m$ are odd. Hence all $a_i$ are even, so $n$ is a power of $2$. Suppose that $n > 8$. We have that $a_1 = 2, a_2 = 4, a_3 = 8$, so $a_1 + 1 = 3, a_2 + 1 = 5, a_3 + 1 = 9$, meaning that $15$ divides $m$ and $m\ne 15$. If $a_t + 1 = 15$ for some $t$, we have $a_t = 14$, contradiction since $14$ isn't a power of $2$. Therefore, $n \le 8$ and since $n$ is composite, $n$ must either be $4$ or $8$.

Wizard_32 wrote: RagvaloD wrote: $n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all the divisors for some $m$ (except $1,m$). Find all such $n$. Assume $k \geq 4$. Hence we get the chains $$n=a_1a_k=a_2a_{k-1}=\cdots$$$$m=(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1)=\cdots \implies u=a_1+a_k=a_2+a_{k-1}=\cdots$$Hence, the roots of the quadratic $x^2-ux+n=0$ are all the pairs $\{a_1,a_k\},\{a_2,a_{k-1}\}, \cdots$ which is a contradiction as $a_1<a_2<\cdots<a_k$. Hence, $k \le 3$. If $k=3$, then $n=a_2^2=a_1a_3$ and so $m=(a_2+1)^2=(a_1+1)(a_3+1) \geq (\sqrt{a_1a_3}+1)^2=(a_2+1)^2$, and thus equality holds and so $a_1=a_3$, a contradiction. Thus, $k=2$ or $1$. It is easy to see that the only possibility here are solutions having $4$ or $3$ distinct positive divisors. Hence, the only possible solutions are $n \in \ \{p^2, p^3, pq\}$ where $p \ne q$ are some primes. But since the aforementioned numbers are the divisors of $m$, hence we rule out some cases. $\bullet$ If $n=p^2$, then $m$ has only $3$ divisors, namely $1,p^2+1,m$ and so $p^2+1$ is a prime. Hence, $p$ must be even, and so $p=2$. Also, when $n=4$, we get $m=25$ which satisfies our conditons. $\bullet$ If $n=p^3$, then the divisors of $m$ are $1, p+1, p^2+1, m$, and so $p+1$ is a prime. Thus, $p=2$ giving $n=8$. In that case, $m=15$ works. $\bullet$ If $n=pq$, then the divisors of $m$ are $1, p+1, q+1, m$, and again we get $p=2$. Since $p \ne q$, $q>3$. But then $q+1$ has a prime divisor that is not $p+1=3$, a contradiction. Hence, the only possible solutions are $\boxed{n \in \{4,8\}}$ Sir I guess you got it a little wrong. Here when u case work for $n = p^2$, $m$ must have factors 1, $p+1$ and $m$. The answer is correct though but you misinterpreted the values of $m$. Btw loving your book

We claim the only answers are $\boxed{n = 4, 8}$ for $\boxed{m = 9, 15}$ respectively. Assume $k \geq 3$. Hence we get the chains, $$n=a_1a_k=a_2a_{k-1}=\cdots$$$$m=(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1)=\cdots$$Hence, we also get $$a_1 + a_k = a_2 + a_{k-1} = \cdots$$Therefore, the sequence $a_1, a_2, \dots, a_k$ is an AP as well as GP. For $k \geq 3$ where the elements are distinct, this is impossible. For $k = 2$ this is only possible when $a_2 - a_1 = \frac{a_2}{a_1}$. Now we claim that $a_2 = 4$ and $a_1 = 2$. After factorising and manipulating the above equation, we get $(a_2-a_1)(a_1-1) = a_1$. Since $a_1-1\mid a_1$, we get $a_1 = 2$ and $a_2 = 4$. In this case we get $n = 8$ and $m = 15$. For $k = 1$, $a_1$ must be a prime and $a_1+1$ must also be a prime. So $a_1 = 2$. In this case we get $n = 4$ and $m=9$. Hence the only two solutions are $\boxed{n = 4, 8}$.

Note that $a_ia_{k-i+1}=n$ and $(a_i+1)(a_{k-i+1}+1)=m$ for all $1\leq i\leq k$. We claim that $n$ has $3$ or $4$ divisors. Suppose FTSOC $n$ has $\geq 5$ divisors. Note that $k\geq 3$. Then $(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1)=m$ so $a_1+a_k=a_2+a_{k-1}$. It follows that $a_1+\frac{n}{a_1}=a_2+\frac{n}{a_2}$ so $a_2-a_1=\frac{n(a_2-a_1)}{a_1a_2}$. Since $a_1\neq a_2$, we have $n=a_1a_2$ so $a_2=a_k$, a contradiction. Hence $n=p^2$ or $n=pq$ for primes $p$ and $q$. In the first case, $m=(p+1)^2$ has $3$ divisors so $p+1$ is prime. This only happens when $p=2$, giving $\boxed{n=4}$. In the second case, $m=(p+1)(q+1)$. Since one of $p$ or $q$ is odd, $2\mid m$. But neither $p+1$ nor $q+1$ is $2$, so there are no solutions in this case. $\square$

First note that $a_1$ is the smallest prime factor of $n$, and $a_1+1$ is the smallest prime factor of $m$. That means both $a_1, a_1+1$ are primes, and the only consecutive primes are $2,3$ so $a_1=2$. If no more divisors exist then $n=4$, and $a_1+1=3 \implies m=9$ which clearly works. If more divisors exist, consider $a_2=p_2$ where $p_2 \neq 2$ or $a_2=4$. If $a_2=4$, then note that $a_{k-1}=\frac{n}{4}$. Then we get that $m=(4+1)(\frac{n}{4}+1)=(3)(\frac{n}{2}+1)$. Manipulating, we get that $n=8$. So we have $1<a_1=2<a_2=4<8$ which clearly works as that generates $1<3<5<15$ for $m=15$. Now consider if $a_2=p_2>2$. Then since $2,3$ are the only consecutive primes, we know that $a_2+1$ must have be composite, thus there must be primes $d, e|a_2+1$. Suppose $d \neq e$, then we do not have enough primes in $1,3$. So $d=e$, but that means $d=e=3$ which means $a_2=8$, contradicting the fact that it is a prime. Thus the solutions are $n=4,8$.

Solution: Only $\boxed{n=4}$ and $\boxed{n=8}$ works. Proof: Claim 1: $k<4$ Proof: FTSOC, Assume $k \geq 4$ We have, $a_1.a_k=a_2.a_{k-1} \iff a_1 = \frac{a_2.a_{k-1}}{a_k} \cdots \cdots eq^n(i)$ We also have, $(a_1+1)(a_k+1)=(a_2+1)(a_{k-1}+1) \iff a_1.a_k+a_1+a_k+1=a_2.a_{k-1}+a_2+a_{k-1}+1 \cdots \cdots eq^n(ii)$ From, $eq^n$(i) and (ii), $a_2.a_{k-1}+a_k^2 = a_2.a_k+a_{k-1}.a_k \iff a_{k-1} = a_k$ which gives a contradiction Claim 2: $k <3$ Proof: FTSOC, Assume $k=3 \implies a_1.a_3=a_2^2$ and $(a_1+1)(a_3+1)=(a_2+1)^2 \iff a_1.a_3+a_1+a_3=a_2^2+2a_2 \implies a_2 = \frac{a_1+a_3}{2}$ Now, $(\frac{a_1+a_3}{2})^2 = a_1.a_3 \iff (a_1-a_3)^2 = 0 \iff a_1=a_3$ which gives a contradicition. Combining this with Claim 1 proves the claim. Case 1: $k =1$ This implies that n is a square of a prime. Let, $ n = a^2 \implies m = (a+1)^2$. If $m$ is even, $2 \mid m$ along with some other number, so it has more amount of divisors than n. So, $m$ must be odd, for m to be odd, a should be even and now the only even prime is $2$. Therefore, $ n = 2^2 = 2$ Upon checking, $m = 9$ works. // Case 2: k = 2 The prime factos are $1,a,b,n$. where $a$ and $b$ are both primes or $b$ is a square of a where a is a prime. If $a$ and $b$ are both prime, $m=(a+1)(b+1) \implies 2 \mid m$ thus increasing the no. of divisors. So, $n$ is a cube. Again, if $n$ is odd, the no. of divisors of $m$ are increased. So, $a$ being an even prime, $n=8$. Upon checking, $m=15$ works. $\blacksquare$

Here is my solution. We claim that the only answers are $\boxed{n=4,8}$ Claim:$n=p^{\alpha}$ Proof: Assume FTSOC that $n$ has at least two distinct prime divisors. Let those smallest prime divisors of $n$ be $p<q$. Note that $a_2=p$ ($\because p$ is the smallest) so $a_2+1=p+1>2$ is the smallest prime divisor of $m$ On the other hand we know that $q+1 \mid m$ also since $q$ is odd we know that $q+1$ is even $\iff 2 \mid q+1$. Hence we get: $2 \mid q+1 \mid m \implies 2 \mid m$ but this contradicts the fact that $p+1>2$ is the smallest prime divisor of $m$. So $n$ has only one prime factor $\iff n=p^{\alpha}$ Claim: $p=2 \iff n=2^{\alpha}$ Proof: FTSOC assume $p>2$ so $p$ odd. Now similarly as before we know that $p+1>2$ is the smallest divisor of $m$ but since $p$ is odd we get that $p+1$ is even so $2 \mid p+1 \mid m$ but this contradicts the fact that $p+1>$ is the smallest divisor of $m$. So our assumption is wrong hence $p=2 \iff n=2^{\alpha}$ $\square$ Claim: $\alpha \leq 3$ Proof: FTSOC assume $\alpha \geq 4$ If $\alpha=4$ then $n=16$ so $1<a_1=2<4=a_2<a_3=8<16=n \therefore \{a_1,a_2,a_3 \}=\{2,4,8 \}\implies\{a_1+1,a_2+1,a_3+1 \}=\{3,5,9 \}$ Since $3 \mid m$ , $5 \mid m \implies 15 \mid m$ so this is possible only if $m=15$ ($\because 3,5,9$ are the only divisors of $m$ except $1,m$) but if $m=15$ then $9$ is not a divisor of $m$ because $9 \nmid 15$ so $\alpha=4$ doesn't work. If $\alpha \geq 5$ then $1<a_1=2<4=a_2<a_3=8<16=a_4 < \dots <n$ so $ \{a_1,a_2,a_3,a_4 \dots \}=\{2,4,8,16 \dots \}\implies\{a_1+1,a_2+1,a_3+1,a_4+1, \dots \}=\{3,5,9,17 \dots \}$ As before since $3 \mid m$ and $5 \mid m$ we get that $15 \mid m$ but $15 $ doesn't appear as a divisor of $m$ above $\rightarrow \leftarrow$ So $\alpha \leq 3$ $\square$ Checking the cases for $\alpha=1,2,3$ we see that $n=4,8$ gives us solutions by picking $m=9,15$ and for $n=2$ we see that it isn't composite $\blacksquare$

One tricky pony aah problem. Replace $a_i$ with $d_i$ because why not. Claim: $d_i+d_{k+1-i}$ is constant. Proof: See that \[d_id_{k+1-i}=n \text{ and }(d_i+1)(d_{k+1-i}+1)=m \implies d_i+d_{k+1-i}=m-n-1\]And done. $\square$. If $n=p^2$, then $p+1$ is the only non trivial factor of some $m$, so $p=2$ and $n=4$. Say $a \neq b$ are smallest factors of $n$. And so \[a+ \frac na=b+\frac nb \implies n=ab\]So $a+1$ and $b+1$ are only non trivial factors of some $m$. So either both are prime and one is square of another and just do some case bash and see that $n=8$ is the only thing which works here. So only solutions are $\boxed{n=4,8}$.