Find all pairs of relatively prime integers $(x, y)$ that satisfy equality $2 (x^3 - x) = 5 (y^3 - y)$.
1. Start with the given equation:
   \[
   2(x^3 - x) = 5(y^3 - y)
   \]
   Rewrite it as:
   \[
   2x^3 - 5y^3 = 2x - 5y
   \]

2. Consider the equation modulo \(2x - 5y\):
   \[
   2x^3 \equiv 5y^3 \pmod{2x - 5y}
   \]
   Multiply both sides by 25:
   \[
   25 \cdot 2x^3 \equiv 25 \cdot 5y^3 \pmod{2x - 5y}
   \]
   Simplify:
   \[
   50x^3 \equiv 125y^3 \pmod{2x - 5y}
   \]
   Since \(125 \equiv 8 \pmod{2x - 5y}\):
   \[
   50x^3 \equiv 8x^3 \pmod{2x - 5y}
   \]
   Therefore:
   \[
   2x - 5y \mid 42x^3
   \]

3. Since \(2x - 5y\) must divide 42, the possible values for \(2x - 5y\) are the divisors of 42:
   \[
   \pm 1, \pm 2, \pm 3, \pm 6, \pm 7, \pm 14, \pm 21, \pm 42
   \]

4. For each possible value of \(k\), solve the Diophantine equation:
   \[
   2x^3 - 5y^3 = 2x - 5y = k
   \]

5. Check each case:
   - For \(k = \pm 1\):
     \[
     2x^3 - 5y^3 = 2x - 5y = \pm 1
     \]
     This implies \(5y \mid 3\), which is a contradiction.

   - For \(k = \pm 3\):
     \[
     2x^3 - 5y^3 = 2x - 5y = \pm 3
     \]
     This implies \(y \mid 3\).

   - For \(k = \pm 7\):
     \[
     2x^3 - 5y^3 = 2x - 5y = \pm 7
     \]
     This implies \(y \mid 63\).

   - For \(k = \pm 21\):
     \[
     2x^3 - 5y^3 = 2x - 5y = \pm 21
     \]
     This implies \(5y \mid 9177\), which is a contradiction.

6. The valid solutions are:
   \[
   (x, y) = (0, \pm 1), (\pm 1, 0), (\pm 4, \pm 3), (\pm 1, \mp 1), (0, 0)
   \]

The final answer is \( \boxed{ (x, y) = (0, \pm 1), (\pm 1, 0), (\pm 4, \pm 3), (\pm 1, \mp 1), (0, 0) } \)