Find all integer triples $(a, b, c)$ satisfying the equation
$$
    5 a^2 + 9 b^2 = 13 c^2.
$$

To find all integer triples \((a, b, c)\) satisfying the equation 
\[ 5a^2 + 9b^2 = 13c^2, \]
we will use properties of quadratic residues and modular arithmetic.

1. **Consider the equation modulo 13:**
   \[ 5a^2 + 9b^2 \equiv 0 \pmod{13}. \]
   This simplifies to:
   \[ 5a^2 \equiv -9b^2 \pmod{13}. \]
   Since \(-9 \equiv 4 \pmod{13}\), we have:
   \[ 5a^2 \equiv 4b^2 \pmod{13}. \]

2. **Analyze the quadratic residues modulo 13:**
   The quadratic residues modulo 13 are the numbers that can be expressed as \(x^2 \pmod{13}\) for some integer \(x\). The quadratic residues modulo 13 are:
   \[ 0, 1, 3, 4, 9, 10, 12. \]

3. **Check if 5 is a quadratic residue modulo 13:**
   We need to determine if 5 is a quadratic residue modulo 13. This can be done using the Legendre symbol:
   \[ \left( \frac{5}{13} \right). \]
   By the law of quadratic reciprocity:
   \[ \left( \frac{5}{13} \right) = \left( \frac{13}{5} \right) \cdot (-1)^{\frac{5-1}{2} \cdot \frac{13-1}{2}}. \]
   Since \(\frac{5-1}{2} = 2\) and \(\frac{13-1}{2} = 6\), we have:
   \[ (-1)^{2 \cdot 6} = (-1)^{12} = 1. \]
   Therefore:
   \[ \left( \frac{5}{13} \right) = \left( \frac{13}{5} \right). \]

4. **Evaluate \(\left( \frac{13}{5} \right)\):**
   Since \(13 \equiv 3 \pmod{5}\), we have:
   \[ \left( \frac{13}{5} \right) = \left( \frac{3}{5} \right). \]
   Using quadratic reciprocity again:
   \[ \left( \frac{3}{5} \right) = \left( \frac{5}{3} \right) \cdot (-1)^{\frac{3-1}{2} \cdot \frac{5-1}{2}}. \]
   Since \(\frac{3-1}{2} = 1\) and \(\frac{5-1}{2} = 2\), we have:
   \[ (-1)^{1 \cdot 2} = (-1)^2 = 1. \]
   Therefore:
   \[ \left( \frac{3}{5} \right) = \left( \frac{5}{3} \right). \]

5. **Evaluate \(\left( \frac{5}{3} \right)\):**
   Since \(5 \equiv 2 \pmod{3}\), we have:
   \[ \left( \frac{5}{3} \right) = \left( \frac{2}{3} \right). \]
   Using quadratic reciprocity again:
   \[ \left( \frac{2}{3} \right) = (-1)^{\frac{3^2-1}{8}} = (-1)^{1} = -1. \]

6. **Conclusion:**
   Since \(\left( \frac{5}{13} \right) = \left( \frac{13}{5} \right) = \left( \frac{3}{5} \right) = \left( \frac{5}{3} \right) = \left( \frac{2}{3} \right) = -1\), we conclude that 5 is not a quadratic residue modulo 13. This means that the equation \(5a^2 \equiv 4b^2 \pmod{13}\) has no non-trivial solutions.

Therefore, the only integer solution to the equation \(5a^2 + 9b^2 = 13c^2\) is \((a, b, c) = (0, 0, 0)\).

The final answer is \( \boxed{ (0, 0, 0) } \).