Let $a$ and $b$ be two positive integers satifying $gcd(a, b) = 1$. Consider a pawn standing on the grid point $(x, y)$. 
A step of type A consists of moving the pawn to one of the following grid points: $(x+a, y+a),(x+a,y-a), (x-a, y + a)$ or $(x - a, y - a)$. 
A step of type B consists of moving the pawn to $(x + b,y + b),(x + b,y - b), (x - b,y + b)$ or $(x - b,y - b)$.
Now put a pawn on $(0, 0)$. You can make a (nite) number of steps, alternatingly of type A and type B, starting with a step of type A. You can make an even or odd number of steps, i.e., the last step could be of either type A or type B. 
Determine the set of all grid points $(x,y)$ that you can reach with such a series of steps.
1. **Reinterpret the Problem Using Gaussian Integers:**
   - Consider the grid points \((x, y)\) as Gaussian integers \(z = x + yi\).
   - A step of type A moves the pawn from \(z\) to \(z + (1 + i)ac\) for some \(c \in \{1, i, -1, -i\}\).
   - A step of type B moves the pawn from \(z\) to \(z + (1 + i)bc\) for some \(c \in \{1, i, -1, -i\}\).

2. **Modulo \(1 + i\):**
   - Each step does not change the value of the point modulo \(1 + i\).
   - Therefore, all reachable points must be divisible by \(1 + i\).

3. **Set of Reachable Points:**
   - Consider the set \(T = \{(1 + i)(ac + bd) : c, d \in \{1, i, -1, -i\}\}\).
   - \(T\) is the set of all Gaussian integers \(t\) such that the pawn can move from \(z\) to \(z + t\) after exactly 2 steps, one of type A and another of type B.

4. **Properties of \(T\):**
   - For any \(c \in \{1, i, -1, -i\}\), \((1 + i)(a + bc) \in T\).
   - For any \(t, u \in T\), \(tc, t + u \in T\) for each \(c \in \{1, i, -1, -i\}\).

5. **Case 1: \(a + b\) is Odd:**
   - We claim that \(1 + i \in T\).
   - Since \(a + b\) is odd, so is \(a - b\), and thus \(\gcd(a + b, a - b) = \gcd(a + b, 2a) = \gcd(a + b, a) = 1\) (in \(\mathbb{Z}\)).
   - Using Bezout's lemma, there exist \(q, r \in \mathbb{Z}\) such that \(q(a + b) + r(a - b) = 1\).
   - Since \((1 + i)(a + b) \in T\) and \((1 + i)(a - b) \in T\), we have \(1 + i = (1 + i)(q(a + b) + r(a - b)) \in T\).

6. **Case 2: \(a\) and \(b\) are Odd:**
   - We claim that \(2 \in T\).
   - In this case, \(\gcd(a + bi, a - bi) = \gcd(a + bi, 2bi) = \gcd(a + bi, 2i) = \{(1 + i)c : c \in \{1, i, -1, -i\}\}\) (in \(\mathbb{Z}[i]\)) as \(\gcd(a + bi, b) = 1\).
   - By Bezout's lemma, there exist \(q, r \in \mathbb{Z}[i]\) such that \(q(a + bi) + r(a - bi) = 1 - i\).
   - Since \((1 + i)(a + bi), (1 + i)(a - bi) \in T\), we find \(2 = (1 + i)(1 - i) \in T\).

7. **Reachability of Points:**
   - Let \(z\) be an arbitrary point divisible by \(1 + i\).
   - If \(2 \mid z\), then \(z \in T\).
   - Otherwise, since \(a\) is odd, we have \(2 \mid z - a(1 + i)\), so \(z - a(1 + i) \in T\).
   - This implies that the pawn can reach \(z - a(1 + i)\) from \(0\) using a valid sequence of alternating steps ending with a step of type B.
   - Then, we finish with the step \(z - a(1 + i) \mapsto z\) of type A.

\(\blacksquare\)

The final answer is any grid points \( \boxed{ (x, y) } \) with \(x + y\) even.