Denote by $\mathbb Z^2$ the set of all points $(x,y)$ in the plane with integer coordinates.  For each integer $n\geq 0$, let $P_n$ be the subset of $\mathbb Z^2$ consisting of the point $(0,0)$ together with all points $(x,y)$ such that $x^2+y^2=2^k$ for some integer $k\leq n$.  Determine, as a function of $n$, the number of four-point subsets of $P_n$ whose elements are the vertices of a square.
1. **Characterize \( P_n \)**:
   - For even \( n = 2k \), the solutions to \( x^2 + y^2 = 2^n \) are \( (\pm 2^k, 0) \) and \( (0, \pm 2^k) \). Let this set of four points be \( B_k \).
   - For odd \( n = 2k+1 \), the solutions to \( x^2 + y^2 = 2^n \) are \( (\pm 2^k, \pm 2^k) \). Let this set of four points be \( R_k \).

   Therefore, the set \( P_n \) can be characterized as:
   \[
   P_n = \{(0,0)\} \cup \left( \bigcup_{i=0}^{\left\lfloor \frac{n-1}{2} \right\rfloor} R_i \right) \cup \left( \bigcup_{i=0}^{\left\lfloor \frac{n}{2} \right\rfloor} B_i \right)
   \]

2. **Prove the characterization by induction**:
   - **Case 1: Even \( n \)**
     - **Base case**: For \( n = 0 \), the only solutions to \( x^2 + y^2 = 2^0 = 1 \) are \( (\pm 1, 0) \) and \( (0, \pm 1) \).
     - **Inductive step**: Assume the statement is true for \( n = 2k \). For \( n = 2(k+1) \), consider \( x^2 + y^2 = 2^{2k+2} \). Since \( x^2 \equiv 0, 1 \pmod{4} \) and \( 2^{2k+2} \) is divisible by 4, \( x \) and \( y \) must be even. Let \( x = 2x' \) and \( y = 2y' \), then:
       \[
       (2x')^2 + (2y')^2 = 2^{2k+2} \implies (x')^2 + (y')^2 = 2^{2k}
       \]
       By the inductive hypothesis, the solutions are \( (x', y') = (\pm 2^k, 0) \) and \( (0, \pm 2^k) \). Thus, the solutions for \( (x, y) \) are \( (\pm 2^{k+1}, 0) \) and \( (0, \pm 2^{k+1}) \).

   - **Case 2: Odd \( n \)**
     - **Base case**: For \( n = 1 \), the only solutions to \( x^2 + y^2 = 2^1 = 2 \) are \( (\pm 1, \pm 1) \).
     - **Inductive step**: Assume the statement is true for \( n = 2k+1 \). For \( n = 2(k+1)+1 \), consider \( x^2 + y^2 = 2^{2k+3} \). Since \( x^2 \equiv 0, 1 \pmod{4} \) and \( 2^{2k+3} \) is divisible by 4, \( x \) and \( y \) must be even. Let \( x = 2x' \) and \( y = 2y' \), then:
       \[
       (2x')^2 + (2y')^2 = 2^{2k+3} \implies (x')^2 + (y')^2 = 2^{2k+1}
       \]
       By the inductive hypothesis, the solutions are \( (x', y') = (\pm 2^k, \pm 2^k) \). Thus, the solutions for \( (x, y) \) are \( (\pm 2^{k+1}, \pm 2^{k+1}) \).

3. **Count the squares in \( P_n \)**:
   - For \( n = 0 \), \( P_0 \) has exactly one square.
   - Incrementing \( n \) introduces exactly five new squares. These are the squares formed by the new points added in \( B_k \) or \( R_k \).

   Therefore, the total number of squares in \( P_n \) is:
   \[
   5n + 1
   \]

The final answer is \( \boxed{ 5n + 1 } \).