, 2021

Let $a$ be the minimum element in $S$. We will show that this $a$ satisfies the condition. Towards contradiction, suppose $b \in S$ be the minimum element in $S$ such that $a \nmid b$. Define $b_0 = b$ and let $b_{i+1}$ be an element in $S$ so that $b_{i+1}^2 \mid a(a+b_i)$ for $i \ge 0$.
Note that if $b_{i+1} < b$, then by the minimality of $b$ we know that $a \mid b_{i+1}$ and hence
$$a^2 \mid b_{i+1}^2 \mid a(a + b_i)$$
and so $a \mid b_i$ and by induction we have $a \mid b_0$ contradict to the assumption. Therefore, $b_i \ge b$ for all $i$. Moreover, we have the $b_{i+1} < \sqrt{b(b+b_i)}$. By induction we can see that $b_n < \varphi b$, where $\varphi = (1 + \sqrt{5})/2$. As a consequence, we know that
$$b^2 \le b_{i+1}^2 \le a(a + b_i) < (1 + \varphi)b < 3b^2,$$
showing that $b_{i+1}^2 = a(a + b_i)$ or $b_{i+1}^2 = a(a + b_i)/2$.
Now let $p$ be a prime so that $v_p(a) > v_p(b)$. If $p \ne 2$, then by induction we have that $v_p(b_i) < v_p(a)$ for all $i$ and
$$v_p(b_{i+1}) = (v_p(a) + v_p(a + b_i))/2 = (v_p(a) + v_p(b_i))/2,$$
therefore $v_p(a) - v_p(b_i)$ decreases strictly as $i$ increases, which is absurd. Hence $p = 2$.
By the same argument, we know that $v_2(b_{i+1}) = \lfloor (v_2(a) + v_p(b_i))/2 \rfloor$. Then $v_p(a) - v_p(b_i)$ decreases strictly unless $v_2(b_i) = v_2(a) - 1$. Therefore, there must be an $N \in \mathbb{N}$ such that $v_2(b_i) = v_2(a) - 1$ for all $i \ge N$. However, this shows that $b_{i+1}^2 = a(a + b_i)/2$ for all $i \ge N$, and so $b_{i+1} < b_i$ for all $i \ge N$. This is a contradiction.

Let $a$ be the minimum element in $S$. We will show that $a$ satisfies the condition. Towards contradiction, suppose $b \in S$ is the minimum element in $S$ such that $a \nmid b$. Let $x \in S$ be an element such that $x^2 \mid a(a+b)$. If $x^2 \ne a(a+b)$, then $x^2 \le a(a+b)/2 < b^2$, so $x < b$, meaning $a \mid x$ and we must have $a \mid b$ similar to the argument in Sol1. So $a(a+b) = x^2$ for some $x \in S$. Similarly $b(a+b) = y^2$ for some $y \in S$. If $\gcd(a,b) = d$ and $a = dA, b = dB$, then $d^2A(A+B) = x^2$, showing that $A = s^2$ for some positive integer $s$, and similarly $B = t^2$. We also have $A+B = l^2$, showing that $s^2+t^2=l^2$, and $x = dsl, y = dtl$.
Now note that $x, y < b\sqrt{2} < 2b$. Therefore if $p \in S$ satisfies that $p^2 \mid a(a+x)$, then either $a(a+x) = p^2, 2p^2$ or $p < b$. The latter case leads to $a|p$ and so $a|b$ by the same argument above, which is a contradiction. Thus $a(a+x) = p^2, 2p^2$.
Now if $q \in S$ satisfies that $q^2 \mid b(b+y)$, then again $b(b+y) = q^2, 2q^2$ or $q < b$.
The latter leads to $a \mid q$, and so
$$d^2 s^4 = a^2 \mid b(b + y) = d^2 t^3 (t + l),$$
and by $\gcd(s,t) = 1$, we have $s^4 \mid t+l$. However we have $(t+l)(l-t) = s^2$, showing that $s=1$, which is impossible. Hence $b(b+y) = q^2, 2q^2$
Now we will show the following lemma that immediately leads to a contradiction.
Lemma There do not exist positive integers $a, b, x, y$, such that $a(a+b) = x^2, b(a+b) = y^2$, and $a(a+x), b(b+y)$ are perfect squares or the double of perfect squares.
Proof. We have already showed that $a = ds^2, b = dt^2, x = dsl, y = dtl$, where $\gcd(s,t) = 1$ and $s^2 + t^2 = l^2$. WLOG, assume $s$ is odd and $t$ is even, then there exist $u, v$ coprime such that $s = u^2 - v^2, t = 2uv$ and $l = u^2 + v^2$. Now note that
$$a(a + x) = d^2 s^3 (s + l) = 2d^2 s^2 u^2 \cdot s,$$
and this shows that $s$ must be a perfect square (because $s$ is odd), say $s = m^2$. Similarly we have
$$b(b + y) = d^2 t^3 (t + l) = 2d^2 t^2 (u + v)^2$$
showing that $t = n^2$ or $2n^2$ for some positive integer $n$. Hence by $s^2 + t^2 = l^2$, we have $m^4 + n^4 = l^2$ or $m^4 + 4n^4 = l^2$. It is well-known (can be proved by infinite descent) that those two Diophantine equations have no nontrivial solutions, which is a contradiction. $\square$