South African Mathematical Olympiad Third Round

We first observe some natural consequences of property (b):
(i) $0 \in A$, since if $x \in A$ is arbitrary, then $0 = x - x \in A$.
(ii) If $x \in A$, then $-x = 0 - x \in A$.
(iii) If $x, y \in A$, then $x + y = x - (-y) \in A$.
(iv) If $x \in A$, then $kx \in A$ for any integer $k$. This follows easily by induction on $k$ (for positive $k$), using (iii). For negative $k$, apply (ii). For $k = 0$, use (i).

So now we know that $k\sqrt{m^2+1} + l\sqrt{n^2+1} \in A$ for all integers $k$ and $l$, and all positive integers $m$ and $n$.
Let us first try to write the integer 1 as a product of two different elements in $A$: We can find (by inspection) positive integers $k, l, m, n$ such that $1 = k^2(m^2+1) - l^2(n^2+1)$, for example, $k=3, m=1, l=1, n=4$. (There are many other possibilities as well.) Hence,
$$\begin{aligned} 1 &= 3^2(1^2 + 1) - 1^2(4^2 + 1) \\ &= (3\sqrt{1^2 + 1} - 1\sqrt{4^2 + 1})(3\sqrt{1^2 + 1} + 1\sqrt{4^2 + 1}), \end{aligned}$$
a product of two different elements of $A$.
If $p$ is an arbitrary integer, then we have
$$\begin{aligned} p = p \cdot 1 &= p \cdot (3\sqrt{1^2+1} - 1\sqrt{4^2+1})(3\sqrt{1^2+1} + 1\sqrt{4^2+1}) \\ &= (3p\sqrt{1^2+1} - p\sqrt{4^2+1})(3\sqrt{1^2+1} + 1\sqrt{4^2+1}), \end{aligned}$$
again a product of two elements of $A$. These two elements of $A$ are indeed different, since if $3p\sqrt{1^2+1} - p\sqrt{4^2+1} = 3\sqrt{1^2+1} + 1\sqrt{4^2+1}$, then we would have $p = 35 + 6\sqrt{34}$, which is not an integer.