We are tasked with finding all integral solutions to the equation:

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

First, let's rewrite the equation and rearrange the terms:

$$a^2b^2 - a^2 - b^2 = c^2.$$

This suggests that $c^2$ must be non-negative, which means $a^2b^2 \ge a^2 + b^2$.

### Case Analysis:

#### Case 1: $a = 0$ or $b = 0$

Without loss of generality, consider $a = 0$. Then the equation becomes:

$$b^2 + c^2 = 0.$$

This implies that $b^2 = 0$ and $c^2 = 0$, hence $b = 0$ and $c = 0$.

Similarly, if $b = 0$, we also get $a = 0$ and $c = 0$.

Thus, one solution is $(a, b, c) = (0, 0, 0)$.

#### Case 2: $a \neq 0$ and $b \neq 0$

Suppose both $a$ and $b$ are non-zero. Since $a^2b^2 \geq a^2 + b^2$, divide both sides by positive $a^2b^2$:

1. Rearrange the equation to $1 \geq \frac{a^2}{b^2} + \frac{b^2}{a^2}$.

2. By the AM-GM inequality, we have:

$$\frac{a^2}{b^2} + \frac{b^2}{a^2} \geq 2.$$

Therefore, the equation $1 \geq 2$ leads to a contradiction.

This contradiction implies there cannot be any non-zero integer solutions for $a \neq 0$ and $b \neq 0$.

### Conclusion:

The only integral solution satisfying the equation $a^2 + b^2 + c^2 = a^2b^2$ is:

$$\boxed{(0, 0, 0)}.$$