A set $A$ is endowed with a binary operation $*$ satisfying the following four conditions:
(1) If $a, b, c$ are elements of $A$, then $a * (b * c) = (a * b) * c$ ,
(2) If $a, b, c$ are elements of $A$ such that $a * c = b *c$, then $a = b$ ,
(3) There exists an element $e$ of $A$ such that $a * e = a$ for all $a$ in $A$, and
(4) If a and b are distinct elements of $A-\{e\}$, then $a^3 * b = b^3 * a^2$, where $x^k = x * x^{k-1}$ for all integers $k \ge 2$ and all $x$ in $A$.
Determine the largest cardinality $A$ may have.

proposed by Bojan Basic, Serbia
To determine the largest cardinality that the set \( A \) can have, we need to analyze the given conditions carefully.

1. **Associativity**: The operation \( * \) is associative, i.e., for all \( a, b, c \in A \),
   \[
   a * (b * c) = (a * b) * c.
   \]

2. **Cancellation Law**: If \( a * c = b * c \), then \( a = b \) for all \( a, b, c \in A \).

3. **Existence of Identity Element**: There exists an element \( e \in A \) such that \( a * e = a \) for all \( a \in A \).

4. **Special Condition**: If \( a \) and \( b \) are distinct elements of \( A - \{e\} \), then
   \[
   a^3 * b = b^3 * a^2,
   \]
   where \( x^k = x * x^{k-1} \) for all integers \( k \ge 2 \) and all \( x \in A \).

Let's analyze these conditions step by step.

### Case 1: \( a^2 = e \)
If \( a^2 = e \), then \( a \) has a cycle of 2. This means \( a * a = e \). 

- Using the special condition (4), let \( b = a \):
  \[
  a^3 * a = a^3 * a^2.
  \]
  Since \( a^3 = a * a^2 = a * e = a \), we have:
  \[
  a * a = a^2 = e.
  \]
  This implies that \( a \) is its own inverse.

### Case 2: \( a^2 \neq e \)
If \( a^2 \neq e \), we need to consider the implications of the special condition (4). Let \( b = a^2 \):
  \[
  a^3 * a^2 = (a^2)^3 * a^2.
  \]
  Simplifying, we get:
  \[
  a^5 = a^8.
  \]
  Since \( a^5 = a^8 \), we can multiply both sides by \( a^{-5} \) (assuming \( a \) has an inverse):
  \[
  e = a^3.
  \]
  This implies that \( a^3 = e \), meaning \( a \) has a cycle of 3.

### Conclusion
From the above cases, we see that every non-identity element \( a \) in \( A \) must either have a cycle of 2 or 3. 

- If \( a \) has a cycle of 2, then \( a^2 = e \).
- If \( a \) has a cycle of 3, then \( a^3 = e \).

We need to check if we can have elements with different cycles in the same set \( A \).

### Mixed Cycles
Suppose \( a \) has a cycle of 2 and \( b \) has a cycle of 3. Then:
  \[
  a^3 * b = b^3 * a^2.
  \]
  Since \( a^3 = e \) and \( b^3 = e \), we get:
  \[
  e * b = e * a^2.
  \]
  This simplifies to:
  \[
  b = a^2.
  \]
  But \( a^2 = e \) for \( a \) with a cycle of 2, so \( b = e \), which contradicts the assumption that \( b \) is a non-identity element.

Thus, all non-identity elements must have the same cycle.

### Maximum Cardinality
If all non-identity elements have a cycle of 3, the set \( A \) can be:
  \[
  A = \{e, a, a^2\}.
  \]
This set has 3 elements and satisfies all the given conditions.

Therefore, the largest cardinality \( A \) may have is \( \boxed{3} \).