Bump!!!!

Nice one! $m=n$ gives $m|a_m$ for any $m$. We write $a_m=mx_m$ Again $m+1 | a_m+a_1 \implies a_m=(m+1)y_m-a_1$ Thus simple solving of linear diophantines give $$a_m=m(m+1)z_m+ma_1$$Thus as $a_m > 0$, we have $z_m \ge \frac{-a_1}{m+1}$ There exists $M_1$ such that for any $m>M_1$, $z_m \ge 0$ Claim: There are finitely many $m_i$'s such that $z_{m_i}=0$. (Proof)Suppose the contrary. Then we have an infinite sequence of positive integers $\{m_i\}_{i \ge 1}$ such that $z_{m_i}=0$. So we must have $m_i+2 | a_{m_i}+a_2$ $\implies m_i + 2 | m_ia_1+a_2$ $\implies m_i+2 | a_2-2a_1$ Size arguments give $a_2-2a_1=0$ which contradicts the problem statement. Again we take $a_m=(m+2)t_m-a_2$ Let $m$ be odd. Solving we get $a_m=m(m+1)(m+2)p_m+m(m+1)(a_1-\frac{a_2}{2})+ma_1$. The above claim also works here(left as an exercise) by setting it up in a similar manner. Thus after some $m > M_2$, we must have $p_m \ge v$ for any positive integer $v$. So $a_m \ge vm^3 + (3v+a_1-\frac{a_2}{2})m^2+(2v+2a_1-\frac{a_2}{2})m$ Choosing a suitable $v$ will give that the coefficients of $m$ and $m^2$ become non negative. Thus after some $M_o$, $a_m \ge m^3$ for odd $m$. Similarly we can show after some $M_e$, $ a_m \ge m^3$. Taking $M=\max (M_o,M_e)$, after $m>M$, $a_m > m^3$. (a stronger form ig is $a_m \ge cm^3$)

sanyalarnab wrote: Again we take $a_m=(m+2)t_m-a_2$ Let $m$ be odd. Solving we get $a_m=m(m+1)(m+2)p_m+m(m+1)(a_1-\frac{a_2}{2})+ma_1$. The above claim also works here(left as an exercise) by setting it up in a similar manner. Thus after some $m > M_2$, we must have $p_m \ge v$ for any positive integer $v$. So $a_m \ge vm^3 + (3v+a_1-\frac{a_2}{2})m^2+(2v+2a_1-\frac{a_2}{2})m$ Choosing a suitable $v$ will give that the coefficients of $m$ and $m^2$ become non negative. Thus after some $M_o$, $a_m \ge m^3$ for odd $m$. Similarly we can show after some $M_e$, $ a_m \ge m^3$. Taking $M=\max (M_o,M_e)$, after $m>M$, $a_m > m^3$. (a stronger form ig is $a_m \ge cm^3$) I don't think you can conclude "$p_m \ge v$ for any positive integer $v$" since $a_n=n^3$ definitely suffices all conditions.

Umm.. my idea was to fix $v$. Then using the logic of the claim, show $p_m$ has to be greater than $v$ after some cutoff. Now i am not saying $p_m=v$ will be achieved for any $v$ but if we get that after some cutoff $p_m$ crosses $v$ our inequality forms.So yeah "$p_m > v$ for any v" is more appropriate

I don't know how to use latex, so I wrote my solution in mathtype and attached them as pictures. sry for the inconvenience. /cry The key idea is to use CRT and pick suitable n in order to form size argument.

Attachments:

@sanyalarnab Can anyone write in detail the way of solving Diophantine equation to reach: a_m=m(m+1)(m+2)p_m+m(m+1)(a_1-\frac{a_2}{2})+ma_1. And how can we conclude that p_m \ge v for any positive integer v.?

Suppose for the sake of contradiction that $a_n < n^3$ for all $n$. Then we can define the following sequence of positive integers: $$b_n = \frac{n^3}{a_n}$$ Since $a_n$ is positive, $b_n$ is also positive. We will show that this sequence satisfies the conditions of the problem, which will lead to a contradiction. First, we have $b_2 = \frac{8}{a_2} \ne 2b_1 = \frac{2}{a_1}$, so $b_2 \ne 2b_1$. Now, let $m$ and $n$ be positive integers. Since $a_m + a_n$ is a divisor of $m + n$, it follows that $\frac{m + n}{a_m + a_n}$ is an integer. Hence, $$\frac{m + n}{b_m + b_n} = \frac{m + n}{\frac{m^3}{a_m} + \frac{n^3}{a_n}} = (a_m + a_n)\left(\frac{m^3}{a_m} + \frac{n^3}{a_n}\right) = (a_m + a_n)\left(\frac{m^2}{a_m} + \frac{n^2}{a_n}\right) \cdot \left(\frac{m}{a_m} + \frac{n}{a_n}\right)$$ Since $a_m + a_n$ is a divisor of $m + n$, it follows that $\frac{m + n}{a_m + a_n}$ is an integer. Similarly, $\frac{m^2}{a_m} + \frac{n^2}{a_n}$ and $\frac{m}{a_m} + \frac{n}{a_n}$ are also integers, so $\frac{m + n}{b_m + b_n}$ is an integer. This shows that the sequence $b_n$ satisfies the conditions of the problem. However, this is a contradiction, since we assumed that the sequence $a_n$ satisfied the conditions of the problem. Therefore, there must exist an integer $M$ such that for all $n > M$, we have $a_n \ge n^3$

$m|a_m$. Let $b_m=\frac{a_m}{m}$. Assume that $m>max\{b_1,b_2\}>|b_1-b_2|$. We have $m+1|b_m-b_1$ and $m+2|b_m-b_2$ by the given divisibility. $b_1 \neq b_2 \implies b_m-b_1, b_m-b_2$ are not both $0$. If one of them is $0$ and the other is not then we have $|b_1-b_2| \ge m+2$ or $|b_2-b_1| \ge m+1$, contradiction. Thus both are not zero. If $m>max\{b_1,b_2\}>|b_1-b_2|$ we have $b_m=(m+1)k+b_1=(m+2)l+b_2 \implies b_m=(m+2)((m+1)c_m-(b_2-b_1))+b_2$ for $c_m \neq 0$. Now let $a_m=c_m m^3+am^2+bm+c$ for $m>max\{b_1,b_2\}>|b_1-b_2|$. If for any $m>max\{b_1,b_2\}>|b_1-b_2|, c_m \ge 2$ we are done ( a cubic with leading coefficient $2$ is eventually bigger than $x^3$). If not then let $c_t=1$. For any $m>2at^2-t+2(b_2-b_1)$, $m+t|a_m+a_t \implies c_m \neq 1$ otherwise we have $m+t | m^3+am^2+bm+c+t^3+at^2+bt+c \implies m+t|2at^2+2(b_2-b_1)$ contradiction. We can take $M=2at^2-t+2(b_2-b_1)$.