Find all triples $(a, b, p)$ of positive integers, where $p$ is a prime number, such that $a^p - b^p = 2013$.
To find all triples \((a, b, p)\) of positive integers, where \(p\) is a prime number, such that \(a^p - b^p = 2013\), we proceed as follows:

1. **Initial Observations and Bounds:**
   - Since \(a^p - b^p = 2013\) and \(a > b\), we start by considering small values of \(a\) and \(b\) to bound \(p\).
   - For \((a, b) = (3, 2)\), we have \(3^p - 2^p\). We need to find \(p\) such that \(3^p - 2^p = 2013\).
   - Calculate for small primes:
     \[
     \begin{aligned}
     &3^2 - 2^2 = 9 - 4 = 5, \\
     &3^3 - 2^3 = 27 - 8 = 19, \\
     &3^5 - 2^5 = 243 - 32 = 211, \\
     &3^7 - 2^7 = 2187 - 128 = 2059.
     \end{aligned}
     \]
   - Since \(3^7 - 2^7 = 2059 > 2013\), \(p\) must be less than 7. Thus, \(p\) could be 2, 3, or 5.

2. **Case Analysis for \(p = 2, 3, 5\):**
   - For \(p = 5\), we check if \(3^5 - 2^5 = 2013\), but \(243 - 32 = 211\), which is too small.
   - For \(p = 3\), we check if \(3^3 - 2^3 = 2013\), but \(27 - 8 = 19\), which is too small.
   - For \(p = 2\), we have \(a^2 - b^2 = 2013\). This can be factored as:
     \[
     (a - b)(a + b) = 2013.
     \]
   - Factorize 2013:
     \[
     2013 = 3 \times 11 \times 61.
     \]
   - Possible pairs \((a - b, a + b)\) are:
     \[
     \begin{aligned}
     &(3, 671), \\
     &(11, 183), \\
     &(33, 61).
     \end{aligned}
     \]

3. **Solving for \((a, b)\):**
   - For \((a - b, a + b) = (3, 671)\):
     \[
     \begin{aligned}
     a - b &= 3, \\
     a + b &= 671.
     \end{aligned}
     \]
     Solving these equations:
     \[
     \begin{aligned}
     2a &= 674 \implies a = 337, \\
     2b &= 668 \implies b = 334.
     \end{aligned}
     \]
     Thus, \((a, b) = (337, 334)\).

   - For \((a - b, a + b) = (11, 183)\):
     \[
     \begin{aligned}
     a - b &= 11, \\
     a + b &= 183.
     \end{aligned}
     \]
     Solving these equations:
     \[
     \begin{aligned}
     2a &= 194 \implies a = 97, \\
     2b &= 172 \implies b = 86.
     \end{aligned}
     \]
     Thus, \((a, b) = (97, 86)\).

   - For \((a - b, a + b) = (33, 61)\):
     \[
     \begin{aligned}
     a - b &= 33, \\
     a + b &= 61.
     \end{aligned}
     \]
     Solving these equations:
     \[
     \begin{aligned}
     2a &= 94 \implies a = 47, \\
     2b &= 28 \implies b = 14.
     \end{aligned}
     \]
     Thus, \((a, b) = (47, 14)\).

4. **Verification:**
   - For \((a, b) = (337, 334)\):
     \[
     337^2 - 334^2 = (337 - 334)(337 + 334) = 3 \times 671 = 2013.
     \]
   - For \((a, b) = (97, 86)\):
     \[
     97^2 - 86^2 = (97 - 86)(97 + 86) = 11 \times 183 = 2013.
     \]
   - For \((a, b) = (47, 14)\):
     \[
     47^2 - 14^2 = (47 - 14)(47 + 14) = 33 \times 61 = 2013.
     \]

Thus, the solutions are \((337, 334, 2)\), \((97, 86, 2)\), and \((47, 14, 2)\).

The final answer is \( \boxed{ (337, 334, 2) } \), \((97, 86, 2)\), and \((47, 14, 2)\).