We begin with a quite standard Lemma If $ a_1,\ldots,a_n$ are integers such that $ \gcd(a_1,\ldots,a_n) = 1$ then any $ m\in\mathbb{Z}$ can be expressed as a linear combination of $ a_1,\ldots,a_n$, i.e. there exist $ \epsilon_1,\ldots,\epsilon_n\in\mathbb{Z}$ such that $ m = \epsilon_1a_1 + \ldots + \epsilon_na_n$. Proof of Lemma: We will use induction. The base case is $ n = 2$. This one is really well-known, but for completeness, here's its proof: Let $ \gcd(a,b) = 1$ and $ m\in\mathbb{Z}$. The numbers $ 0,a,2a,\ldots,(b - 1)a$ are all different $ \pmod b$. This means one of them, say $ a\epsilon_1$ is $ 1\pmod b$. Then $ 1 - ax = b\epsilon_2$ for some $ \epsilon_2$, hence $ a\epsilon_1 + b\epsilon_2 = 1$. Then $ m = amx + bm\epsilon_2$. Now the induction step. Assume we're done for $ n - 1\ge2$. Let $ d = \gcd(a_1,\ldots,a_{n - 1})$. Since $ \cgd(d,a_n) = 1$ then any integer $ m$ is representable as $ m = d\epsilon + a_n\epsilon_n$. Consider now the numbers $ b_i = \dfrac{a_i}d$, for $ i = \overline{1,n - 1}$. Then $ \gcd(b_1,\ldots,b_{n - 1}) = 1$ and according to the Induction Hypothesis, $ 1 = \alpha_1b_1 + \ldots + \alpha_{n - 1}b_{n - 1}$, for some $ \alpha_1,\ldots,\alpha_{n - 1}\in\mathbb{Z}$. Then $ d = \alpha_1a_1 + \ldots + \alpha_{n - 1}a_{n - 1}$, and therefore $ m = \epsilon d + a_n\epsilon_n = \epsilon_1a_1 + \ldots + \epsilon_na_n$, where $ \epsilon_i = \epsilon\alpha_i$, for $ i = \overline{1,n - 1}$. ___ Now back to the problem. Take 3 elements $ a,b,c$ of $ S$ (it's trivial that $ S$ has at least 3 elements). Then $ b^2 - a\in S$, and $ c^2 - (b^2 - a) = a + (c^2 - b^2)\in S$. We prove by induction on $ k$ that $ a + k(c^2 - b^2)\in S$. Base case $ k = 1$ is done. Assume the statement true for $ k$. Then $ b^2 - [a + k(c^2 - b^2)]\in S$, and consequently $ c^2 - b^2 + a + k(c^2 - b^2)\in S$. Changing $ b,c$ with places we obtain the following result: $ a + n(c^2 - b^2)\in S$, for all $ n\in\mathbb{Z}$. Immediately follows that $ a + n_1c_1 + \ldots + n_tc_t$ are in $ S$ for all $ n_i\in\mathbb{Z}$, where $ c_i$ are of the form $ x^2 - y^2$ with $ x,y\in S$. We are left to show that there is a set of numbers of the form $ c_i = x^2 - y^2$ with $ x,y\in S$ such that $ \gcd(c_1,\ldots,c_?) = 1$ (after that we just apply our Lemma). Let $ M = \{x^2 - y^2 > 0|x,y\in S\} = \{m_0,m_1,\ldots\}$. Let $ b_i = \gcd(m_0,\ldots,m_i)$. It is clear that $ b_i$ is decreasing. If $ b_i > 1$ for all $ i$, then from some point on it's constant, thus there exists a prime number $ p$ dividing all positive differences $ x^2 - y^2$, for $ x,y\in S$. We show that this is impossible. Suppose $ p|x^2 - y^2$, for all $ x,y\in S$. Take $ a,b\in S$ such that $ \gcd(a,b) = \gcd(a - 2,b - 2) = 1$. WlOG, $ p\not|a$. Since $ p|a^2 - b^2$ we get $ p\not|b$. Also $ p|(a^2 - a)^2 - b^2$ and we have $ p|a^3(a - 2)$, thus $ p|a - 2$. Similarly $ p|b - 2$, hence $ \gcd(a - 2,b - 2)\neq1$, contradiction.