misread sorry

vsathiam wrote: Now let the smallest element of M be x. From the problem statement, all numbers of the form kx belong in M, for all natural numbers k. Why must $x \in M$ imply $kx \in M$ for all $k \ge 1$? That is not implied by the problem statement.

It exists We just have to construct it inductively

Let $\sigma (n)$ be the sum of the divisors of $n$ and $f(n) = \dfrac{\sigma (n)}{n}$; we're interested in a set $M$ with $f(a+b) <2$ for all $a,b\in M$. Let $M_1 = \{ 1\}$. We'll construct a sequence of sets $M_1,M_2,...$ with $M_{k+1} = M_k \cup a_{k+1}$ for some positive integer $a_{k+1}$ with $a_1=1$; taking the union of all these sets will produce an infinite set $M$. Let $p_1<p_2<...$ be the primes in increasing order. Suppose at some step we have the set $M_k = \{a_1,a_2,...,a_k\}$. We'll choose $a_{k+1} = p_1p_2...p_m+1$ for sufficiently large $m$ to be determined later. Clearly we just need $a_i+a_{k+1}$ to be good for each $i\le k+1$. Lemma: For all $n$, $f(n)< \displaystyle\prod_{p|n} \left( \dfrac{p}{p-1} \right)$. Proof: Note $f(n) = \displaystyle\prod_{p|n} \left( 1 + p^{-1} + ... + p^{-v_p (n)}\right) < \displaystyle\prod_{p|n} \left( \dfrac{p}{p-1}\right)$ by the infinite geometric series formula applied to each term of the product. First we'll show that for all large $m$, our choice of $a_{k+1}$ satisfies that $a_{k+1}+a_{k+1}=2a_{k+1}$ is good. Indeed, all primes dividing $a_{k+1}$ must have size at least $p_{m+1}$, hence $f(2a_{k+1}) = 1.5f(a_{k+1}) < 1.5\displaystyle\prod_{p|a_{k+1}} \left( \dfrac{p}{p-1} \right) \le 1.5\displaystyle\prod_{p|a_{k+1}}\left( \dfrac{p_{m+1}}{p_{m+1}-1} \right) \le \dfrac{3}{2} \cdot \left( \dfrac{p_{m+1}}{p_{m+1}-1} \right)^{\log_{p_{m+1}} a_{k+1} }$. But it's not hard to see that $a_{k+1} = p_1p_2...p_m + 1 \le p_{m+1}^m$, hence this last term is at most $1.5 \cdot \left( 1 + \dfrac{1}{p_{m+1}-1} \right)^m$. For large enough $m$, the ratio $b_m= \dfrac{m}{p_{m+1}-1}$ grows arbitrarily small (as roughly $(\log m)^{-1}$), hence $1.5 \cdot \left( 1 + \dfrac{1}{p_{m+1}-1} \right)^m \approx 1.5 \cdot e^{b_m} < 2$, as desired. Next, we'll show that for all sufficiently large $m$, each $a_i + a_{k+1}$ must be good. We'll assume that $p_m > a_k$. Then proof of this claim is similar- we write $a_i +a_{k+1} = (a_i + 1) + p_1p_2...p_m$. By the inductive hypothesis, $a_i+1=a_i+a_1$ is good, hence $f(a_i+1) = c_i <2$. Since $p_m \ge a_k+1 \ge a_i+1$, if $p\not | a_i+1$ and $p|a_i+a_{k+1}$, we know $p\not | p_1p_2...p_m \implies p \ge p_{m+1}$. It follows that $f(a_i+a_{k+1}) < \displaystyle\prod_{p|a_i+a_{k+1}, p |a_i+1} \left( \dfrac{p}{p-1} \right) \displaystyle\prod_{p|a_i+a_{k+1}, p \ge p_{m+1}} \left( \dfrac{p}{p-1} \right) \le c_i \displaystyle\prod_{p|a_i+a_{k+1}, p \ge p_{m+1}} \left( \dfrac{p}{p-1} \right) \le c_i \left( \dfrac{p_{m+1}}{p_{m+1}-1} \right)^{\log_{p_{m+1}} (a_i + a_{k+1} ) }$. Then for each $a_i+a_{k+1}$ to be good, we just need $\left( \dfrac{p_{m+1}}{p_{m+1}-1} \right)^{\log_{p_{m+1}} (a_i + a_{k+1} ) }\le \dfrac{2}{c_i}$ for each $i$, where each $c_i$ is less than $2$. But again, $a_i + a_{k+1} < 2(p_1p_2...p_m+1) < 4p_2p_3...p_m\le p_{m+1}^m$, hence the LHS of the previous inequality is at most $\left( \dfrac{p_{m+1}}{p_{m+1} -1} \right)^m$, As with the previous argument, for large $m$ this term approaches $e^{b_m}$, which grows arbitrarily close to one as $b_m \to 0$, hence for suitable $m$ we have $LHS \le \text{min} \left( \dfrac{2}{c_1}, ..., \dfrac{2}{c_k} \right)$, hence each $a_i+a_{k+1}$ is good, as desired.

Wait @Mod can someone pls take down my post/fake solve? i think its a misread...