Let $a$ be the smallest element in $S$, and FTSOC assume some $b \in S$ is not divisible by $a$. Choose the smallest such $b$, and let $c^2$ divide $a(a+b)$. If $c<b$, then $$a^2 \mid c^2 \mid a(a+b)$$$\implies$ $a \mid b$, contradiction! So $c \geq b$. Then it is easy to get $b<2a$. Now define a sequence $b_1,b_2, \dots$ satisfying $b_1=b$ and $b_{i+1} \in S$ such that $b_{i+1}^2 \mid a(a+b_i)$ for all $i \geq 1$ (for multiple such values, choose the smallest possible $b_i$. It is easy to prove using induction that $a \nmid b_i$ and $b_i <2a$ for all $i$. Therefore the sequence $b_i$ is bounded, and so must be periodic. Rename the repeating unit as $r_1, \dots r_k$. This will be sufficient to obtain a contradiction. First, divide through $a,r_1,\dots r_k$ by their common divisor (this can be done since all divisibilities are homogenous). Assume the reduced values from now on. Clearly $a>1$ because of our initial assumption. Now, $a(a+r_i) \geq 3r_{i+1}^2$ $\implies$ $ar_i > 2r_{i+1}^2 >2a^2$, which contradicts $r_i <2a$. Therefore $$a(a+r_i)=e_ir_{i+1}^2$$for all $i$ where $e_i$ is $1$ or $2$. Now choose any prime $p$ and let $m=v_p(\gcd(a,r_i))$. Then $$2m \leq v_p(a(a+r_i)) \leq 1 + 2v_p(r_{i+1})$$since $e_i$ is either $1$ or prime. So $v_p(r_{i+1}) \geq m$, and so doing this for all primes, we get $\gcd(a,r_{i+1}) \geq \gcd(a,r_i)$. But since the sequence is cyclic, all gcds have the same value, say $d$. Since all of them are coprime, we must have $d=1$. But then, from the original relation, we must have $a \leq e_i \leq 2$, and so $r_{i+1}^2 \mid a+r_i$ $\implies$ $r_{i+1}^2 \leq 2+r_i < r_i^2$ (which must hold for $r_i \geq a+1 \geq 3$). But this clearly cannot hold because of cyclicity of $r_i$, contradiction! $\blacksquare$

Any set having a common divisor, reduced by dividing all the terms in the set by the highest common divisor. Let the set $S , $ has the highest common factor of $x $. Note, that we write that as $xa_1,xa_2. $ For any two element from the set, $xa_i ,xa_j $ say $(xa_k)^2\mid xa_i(xa_i+xa_j)\equiv x^2\times a_k^2\mid x^2\times a_i\times (a_i+a_j)\equiv a_k^2\mid a_i (a_i+a_j). $ Set $S $ satisfies the condition $\equiv $ $\frac {S}{x} $ does.

Here is another proof that only looks at finitely many terms. The solution is thus a bit more involved. Let $a$ be the minimum element in $S$ and $b\in S$ be the minimum element in $S$ such that $a\nmid b$. Let $x\in S$ be an element such that $x^2|a(a+b)$. If $x^2\neq a(a+b)$, then $x^2\leq a(a+b)/2<b^2$ (here we use $a<b$), showing that \[a^2\mid x^2\mid a(a+b),\Rightarrow a\mid a+b,\]showing that $a\mid b$, which is a contradiction. Therefore $a(a+b)=x^2$ 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 leads to $a\mid p$ and thus $a\mid x$ with the same argument above, and so $a\mid b$ by the same argument, which is again a contradiction. Hence $a(a+x)=p^2,2p^2$. Now if $q\in S$ satisfies that $q^2\mid b(b+y)$, then again either $b(b+y)=q^2,2q^2$ or $q<b$. The latter leads to $a\mid q$, and so \[d^2s^4=a^2\mid b(b+y)=d^2t^3(t+l),\]and by $\gcd(s,t)=1$ we have that \[s^4\mid t+l.\]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 double some perfect squares.

This entire argument shows that the existence of $b$ will eventually lead to a contradiction, and so there should not exist any $b\in S$ such that $a\nmid b$. This proves the desired statement.

Let the minimal element be $a$. Choose the smallest $x_1\in S$ such that $a\nmid x_1$. Inductively, let $x_{i+1}$ be the minimal integer such that $$x_{i+1}^2\mid a(a+x_i)$$Notice that $a\nmid x_{i+1}$, otherwise $$a^2\mid x_{i+1}^2\mid a(a+x_i)\implies a^2\mid ax_i\implies a\mid x_i$$and inductively, $a \mid x_{i-1} \implies \cdots \implies a\mid x_1$, a contradiction. We also have that if $x_2<x_1$, we get that $a\mid x_2$ by $x_1$'s minimality, so $$a^2\mid x_2^2\mid a(a+x_1)\implies a\mid x_1$$a contradiction. Claim 1: $x_i<2a$ for all $i$. Proof: Since $x_2\geq x_1$ we have $x_1^2\leq x_2^2\leq a(a+x_1)$ so if $x_1=ca$ we have that $$c^2a^2\leq a^2+ca^2\implies c^2-c-1\leq 0$$This gives $c<2$ so $x_1\leq 2a$. Inductively, $$x_{i+1}^2\leq a(a+x_i)<3a^2\implies x_{i+1}<\sqrt{3}a<2a$$so $x_i<2a$ for all $i$. Claim 2: $x_i\neq x_{i+1}$. Proof: FTSOC suppose $x_i=x_{i+1}$ for some $i$. Thus, $$x_i^2\mid a(a+x_i)<x_i(x_i+x_i)=2x_i^2\implies x_i^2=a^2+ax_i$$But since $a\nmid x_i$, choose prime $p$ such that $\nu_p(x_i)<\nu_p(a)$. Then, $$\nu_p(x_i^2)=2\nu_p(x_i)<\nu_p(x_i)+\nu_p(a)=\nu_p(a(a+x_i))=\nu_p(a^2+ax_i)$$a contradiction. Claim 3: $a(a+x_i)\leq 2x_{i+1}^2$. Proof: Suppose the contrary, so $a(a+x_i)\geq 3x_{i+1}^2$ But then, $$3a^2=a(a+2a)>a(a+x_i)\geq 3x_{i+1}^2$$so $a>x_{i+1}$, a contradiction since $a$ is the smallest element in $S$. Claim 4: If $x_i>x_{i-1}$, we have a contradiction. Proof: First, such $i$ must exists because $x_i$ cannot be non-decreasing as it is lowerbounded and takes finitely many values and $x_i\neq x_{i+1}$ for all $i$. Now, let $u^2=a(a+x_i)$. We establish that if $x_i>x_{i-1}$, since $a(a+x_{i-1})\leq 2x_i^2$: \begin{align*} u^2&=a(a+x_i)\\ &<a\left(a+\sqrt{\frac{a(a+x_{i-1})}{2}}\right)\\ &<a^2+\frac{a}{\sqrt{2}}u\\ \implies u^2-\frac{a}{\sqrt{2}}u-a^2&<0\\ u^2&<2a^2\\ a(a+x_i)&<2a^2\\ x_i&<a \end{align*}a contradiction, concluding the proof.

Let $a_1 < a_2 < a_3 < \cdots $ be the elements of $S$ in increasing order. We aim to show that $a_1$ divides every element of $S$. We have for any positive integers $i,j$, there exists a positive integer $k$ so that\[ a_k^2 \mid a_i (a_i + a_j)\] Suppose this was not the case. Fix $b_1, b_2, \ldots, $ so that $b_1$ is the smallest positive integer with $a_1 \nmid a_{b_1}$, and $b_{i + 1} $ is the smallest positive integer so that\[a_{b_{i + 1}}^2 \mid a_1(a_1 + a_{b_{i}})\] Now define $c = a_1$. Note that $a_i > c$ for all $i > 1$. Claim: $c \nmid a_{b_i}$ for each $i$ Proof: Note that if $c \mid a_{b_i} $, then $c^2 \mid c(c + a_{b_{i- 1}})$, so $c \mid a_{b_{i - 1}}$. Repeating this process, we find that $c \mid a_{b_1} $, which is not true. $\square$ Claim: $a_{b_i} \le 2c$ for each $i$ Proof: This is done by induction on $i$. Firstly, we have $a_{b_2}^2 \mid c(c + a_{b_1})$. Note that since $c \nmid a_{b_2}$, we have $a_{b_2} \ge a_{b_1}$ due to the minimality of choice of $a_{b_1}$. However, $c(c + a_{b_1})$ is less than $2a_{b_1}^2 \le 2 a_{b_2}^2$, meaning $c(c + a_{b_1}) = a_{b_2}^2$. Thus, $a_{b_1}^2 \le a_{b_2}^2 = c(c + a_{b_1})$. This gives that $-a_{b_1}^2 + c a_{b_1} + c^2 \ge 0$. Suppose that $a_{b_1} = r \cdot c$. We have $c^2 \cdot r^2 \le c(c + r \cdot c)$, so $r^2 \le r + 1$, which implies $r < 2$, which proves the base case. Now, note that if $a_{b_i} \le 2c$, then $a_{b_{i + 1}}^2 \le c(c + 2c) = 3c^2$, so $a_{b_{i + 1}} \le 2c$ also. $\square$ Thus, $(b_i)$ must be bounded. Since $b_{i +1}$ is uniquely determined from $b_i$, $(b_i)$ must also be eventually periodic. Claim: $a_{b_i} > c$ for each $i$ Proof: Follows from the fact that $c \mid a_{b_i}$ (the idea being that $a_{b_1} > c$ is the smallest number in $S$ that isn't a multiple of $c$). $\square$ Suppose $a_{b_i}$ was eventually periodic with period $d$. Now fix $c_0 = a_{b_N}$ for some $N$ so that $a_{b_i} = a_{b_{i + d}}$ for all $i \ge N$. Now define $c_i = a_{b_{i + N}}$ for all integers $i \ge 0$. Notice that we still have\[ c_{i + 1}^2 \mid c(c + c_i)\]$c < c_i < 2c$, and $c \nmid c_i$ for each integer $i \ge 0$. Also note that $c_i = c_{i + d}$ for each $i$. By size, $\frac{c(c + c_i)}{c_{i + 1}^2} \in \{1,2\}$. Hence for any odd prime $p$,\[ 2\nu_p(c_{i + 1}) = \nu_p(c) + \nu_p(c + c_i) \]Claim: For any odd prime $p$, if $\nu_p(c_i) < \nu_p(c)$, then $\nu_p(c) > \nu_p(c_{i + 1} ) > \nu_p(c_i)$. Proof: The idea is that $\nu_p(c + c_i)$ becomes $\nu_p(c_i)$. Hence $\nu_p(c_{i + 1}) = \frac{\nu_p(c) + \nu_p(c_i)}{2} > \nu_p(c_i)$. $\square$ Now for some odd prime $p$, if $\nu_p(c_i) < \nu_p(c)$, then $\nu_p(c_i) < \nu_p(c_{i + 1} ) < \nu_p(c_{i + 2} ) < \cdots \nu_p(c_{i + d})$, a contradiction since $\nu_p(c_i) = \nu_p(c_{i + d})$. Therefore, $\nu_p(c_i) \ge \nu_p(c)$ for all odd primes $p$. Since $c \nmid c_i$, we have that $\nu_2(c_i) < \nu_2(c)$ Now note that $2\nu_2(c_{i + 1}) = \nu_2(c) + \nu_2(c + c_i)$ or $2\nu_2(c_{i + 1}) = \nu_2(c) + \nu_2(c + c_i) - 1$. However, we have that $\nu_2(c + c_i) = \nu_2(c_i)$. We have $2 \nu_2(c_{i + 1}) \ge \nu_2(c) + \nu_2(c_i) - 1 \ge 2 \nu_2(c_i)$, so $\nu_2(c_{i + 1}) \ge \nu_2(c_i)$. Since $\nu_2(c_i) = \nu_2(c_{i + d})$, $\nu_2(c_i)$ must be constant over all $i$. However, if $2\nu_2(c_{i + 1}) = \nu_2(c) + \nu_2(c_i)$ holds only if $\nu_2(c_{i + 1} ) > \nu_2(c_i)$, which is false. Hence $2\nu_2(c_{i + 1}) = \nu_2(c) + \nu_2(c_i) - 1$ must hold for each $i$, so $2 c_{i + 1}^2 = c(c + c_i)$. However, considering the maximal $c_i$ gives a contradiction, so we are done and $a_1$ must divide every element of the sequence.