Let $S = \mathbb Z \times \mathbb Z$. A subset $P$ of $S$ is called [i]nice[/i] if
[list]
[*] $(a, b) \in P \implies (b, a) \in P$
[*] $(a, b)$, $(c, d)\in P \implies (a + c, b - d) \in P$
[/list]
Find all $(p, q) \in S$ so that if $(p, q) \in P$ for some [i]nice[/i] set $P$ then $P = S$.
1. **Initial Claim and Setup**:
   We claim that the pairs \((p, q)\) which work are those where \(\gcd(p, q) = 1\) and \(p \not\equiv q \pmod{2}\). To prove this, we need to show that if \((p, q)\) is in a nice set \(P\), then \(P = \mathbb{Z} \times \mathbb{Z}\).

2. **Non-working Pairs**:
   If \(\gcd(p, q) = d \neq 1\), then all pairs \((x, y)\) that can be produced from \((p, q)\) will have \(x = dx'\) and \(y = dy'\) for some integers \(x'\) and \(y'\). This means that all elements in \(P\) will be multiples of \(d\), and thus \(P \neq \mathbb{Z} \times \mathbb{Z}\).

3. **Generating Specific Pairs**:
   From \((p, q)\), we can generate \((2p, 0)\) and \((2q, 0)\) as follows:
   \[
   (p, q), (p, q) \implies (2p, 0)
   \]
   \[
   (q, p), (q, p) \implies (2q, 0)
   \]
   Since \((a, b) \in P \implies (b, a) \in P\), we also have \((0, 2p)\) and \((0, 2q)\) in \(P\).

4. **Using the Euclidean Algorithm**:
   Using the Euclidean Algorithm, we can obtain \((0, 2)\) from \((0, 2p)\) and \((0, 2q)\) since \(\gcd(2p, 2q) = 2\).

5. **Parity Considerations**:
   Both \(p\) and \(q\) cannot be odd, as this would only produce pairs of the form (odd, odd) and (even, even). Therefore, one of \(p\) and \(q\) must be even. Without loss of generality, assume \(p\) is even and \(q\) is odd.

6. **Generating \((1, 0)\)**:
   Using the second operation repeatedly:
   \[
   (p, q), (0, 2) \implies (p, 1)
   \]
   \[
   (p, 1) \implies (1, p), (0, 2) \implies (1, 0)
   \]
   Thus, we have obtained \((1, 0)\).

7. **Generating \((0, 1)\) and \((0, -1)\)**:
   From \((0, 1)\) and \((0, 2)\), we can also obtain \((0, -1)\).

8. **Generating All Pairs**:
   From \((p, q)\), we can reach all \((p, x)\) for any integer \(x\):
   \[
   (p, q), (0, 1) \implies (p, q - 1) \implies \dots \implies (p, q - k)
   \]
   \[
   (p, q), (0, -1) \implies (p, q + 1) \implies \dots \implies (p, q + k)
   \]
   Similarly, we can reach any \(p = y\):
   \[
   (p, x) \implies (x, p), (0, 1) \implies (x, p - 1) \implies \dots \implies (x, p - l) \implies (p - l, x)
   \]
   \[
   (p, x) \implies (x, p), (0, -1) \implies (x, p + 1) \implies \dots \implies (x, p + l) \implies (p + l, x)
   \]
   Thus, any value \((y, x)\) can be achieved, and we are done.

\(\blacksquare\)

The final answer is \( \boxed{ (p, q) } \) such that \(\gcd(p, q) = 1\) and \(p \not\equiv q \pmod{2}\).