We are tasked with finding all triples $(a, b, c)$ of integers that satisfy the following equations:

$$a + b = c$$
$$a^2 + b^3 = c^2$$

First, substitute $c = a + b$ from the first equation into the second equation:

$$a^2 + b^3 = (a + b)^2$$

Expanding $(a + b)^2$, we have:

$$a^2 + b^3 = a^2 + 2ab + b^2$$

Subtract $a^2$ from both sides:

$$b^3 = 2ab + b^2$$

Rearrange the terms:

$$b^3 - b^2 - 2ab = 0$$

Factor the left-hand side:

$$b(b^2 - b - 2a) = 0$$

From the factorization, we get two cases to consider:

### Case 1: $b = 0$
- Substitute $b = 0$ into the first equation $a + b = c$, we get:
$$a + 0 = c \quad \Rightarrow \quad c = a$$
- Thus, one solution is:
$$(a, b, c) = (a, 0, a)$$

### Case 2: $b^2 - b - 2a = 0$
- Solve this quadratic equation for $a$:
$$b^2 - b - 2a = 0 \quad \Rightarrow \quad 2a = b^2 - b \quad \Rightarrow \quad a = \frac{b^2 - b}{2}$$

- Substitute $a = \frac{b^2 - b}{2}$ back into the first equation $a + b = c$:
$$\frac{b^2 - b}{2} + b = c$$
$$\frac{b^2 - b + 2b}{2} = c \quad \Rightarrow \quad \frac{b^2 + b}{2} = c$$
- Hence, another solution is:
$$(a, b, c) = \left( \frac{b^2 - b}{2}, b, \frac{b^2 + b}{2} \right)$$

Thus, the solutions for the integer triples $(a, b, c)$ are:
$$\boxed{(a, 0, a) \text{ or } \left( \frac{b^2 - b}{2}, b, \frac{b^2 + b}{2} \right)}$$