Let $c>a>b$ is three minimal numbers in $B$ and $a=xd,b=yd,(x,y)=1$ Then $\frac{a-b}{(a,b)}=b$ $a-b=(a,b)b$ $xd-yd=yd^2 \to x=y(d+1) \to y=1,x=d+1 \to b=d,a=d^2+d$ $\frac{c-b}{(c,b)}=a$ or $b$ But If $\frac{c-b}{(c,b)}=b \to c=a=b^2+b$ So $c-b=(c,b)a \to d|c \to c=zd \to zd-d=d^2(d+1) \to z=d^2+d+1 \to c=d(d^2+d+1)$ $\frac{c-a}{(c,a)}=\frac{d^3}{d}=d^2<a \to d^2=b=d=1$ So for every $x \in B, x-1 \in B$

Let $f(a, b) = \frac{a - b}{(a, b)}$. If $1 \in B$, then for any $a$ in $B$, $f(a, 1) = a-1 \in B \implies a-2 \in B \implies \cdots$ and since there are infinitely many numbers in $B$, if $1 \in B$, then every positive integer must be in $B$. So, it is sufficient to prove that $1 \in B$. Starting from any pair $(a, b)$, we let $f(a, b) = k_1 < a$ (since $b$ is positive). Now we have a new pair $(\text{max}(k_1, b), \text{min}(k_1, b))$, which means $f(\text{max}(k_1, b), \text{min}(k_1, b))= k_2 < \text{max}(k_1, b) \in B$. Then we take the two smallest pairs in $(k_1, k_2, b)$ and continue this process. Continuing like this, we get a strictly decreasing sequence of pairs and hence, at some points, it must reach $1$. So, we are done.

green_leaf wrote: Let $f(a, b) = \frac{a - b}{(a, b)}$. If $1 \in B$, then for any $a$ in $B$, $f(a, 1) = a-1 \in B \implies a-2 \in B \implies \cdots$ and since there are infinitely many numbers in $B$, if $1 \in B$, then every positive integer must be in $B$. So, it is sufficient to prove that $1 \in B$. Starting from any pair $(a, b)$, we let $f(a, b) = k_1 < a$ (since $b$ is positive). Now we have a new pair $(\text{max}(k_1, b), \text{min}(k_1, b))$, which means $f(\text{max}(k_1, b), \text{min}(k_1, b))= k_2 < \text{max}(k_1, b) \in B$. Then we take the two smallest pairs in $(k_1, k_2, b)$ and continue this process. Continuing like this, we get a strictly decreasing sequence of pairs and hence, at some points, it must reach $1$. So, we are done. why if $1 \in B$, then $B$ contains all positive integers?

Jjesus wrote: why if $1 \in B$, then $B$ contains all positive integers? Huh? It is explained in the first paragraph of the solution cited by you: Take any $n$. Then since $B$ is infinite, there is some $m>n$ which is in $B$. But then as shown there, also $m-1 \in B, m-2 \in B$ etc. and hence finally $n \in B$.

Here's a greedy solution. Consider three smallest integers in set $B$, $a_1<a_2<a_3$, I contend, that $a_i=i$ for $i=1,2,3$. We must have $\frac{a_2-a_1}{\gcd{(a_1,a_2)}}\leq a_2-a_1<a_2$, thus $a_2-a_1=a_1\gcd{(a_1,a_2)}$. Now here $a_1\mid a_2$, let $a_2=a_1k$, thus $a_1=k-1$ and therefore $a_2=k(k-1)$. Now, consider the condition with $a_3>a_1$, thus $\frac{a_3-a_1}{\gcd{(a_1,a_3)}}\leq a_3-a_1<a_3$, note that if $\frac{a_3-a_1}{\gcd{(a_1,a_3)}}=a_1$, then we get that $a_2=a_3$, absurd, thus $a_3-a_1=a_2\gcd{(a_1,a_3)}$. Furthermore $a_1\mid a_3$, let $a_3=la_1=l(k-1)$, thus $l-1=k(k-1)\implies a_3=(k-1)(k^2-k+1)$. On the other hand the condition with $a_3>a_2$ implies $\frac{a_3-a_2}{\gcd{(a_2,a_3)}}=\frac{(k-1)^3}{k-1}=(k-1)^2\in B$ as $k>1$. Note that either $(k-1)^2=k(k-1)$ or $(k-1)^2=k-1$. As the former is impossible, we obtain that $k=2$ from the latter. This implies that $a_i=i$ for $i=1,2,3$. Now as $B$ is infinite and $1\in B$, we take can take $b_0\to \infty \in B$ and get $\frac{b_0-1}{\gcd{(b_0,1)}}=b_0-1$ is in $B$, repeat. We are done.

We uploaded our solution https://calimath.org/pdf/BalticWay2018-19.pdf on youtube https://youtu.be/zot_IEIOwzo.