To solve the given system of equations:
$$a^2 = bc + 1$$
$$b^2 = ac + 1$$

we will proceed step-by-step.

1. Subtract the two equations:
$$a^2 - b^2 = bc - ac$$
Using the difference of squares on the left-hand side:
$$(a - b)(a + b) = c(b - a)$$
Simplifying, we get:
$$(a - b)(a + b + c) = 0$$

2. Consider the cases:
- **Case 1: $a - b = 0$**
$$a = b$$
Substitute $b = a$ into the first equation:
$$a^2 = ac + 1$$
$$a^2 - ac - 1 = 0$$
This is a quadratic equation in $a$. Solving for $a$ using the quadratic formula:
$$a = \frac{c \pm \sqrt{c^2 + 4}}{2}$$
Since $a$ must be an integer, $c^2 + 4$ must be a perfect square. Let $k^2 = c^2 + 4$, then:
$$k^2 - c^2 = 4$$
$$(k - c)(k + c) = 4$$
The integer pairs $(k - c, k + c)$ that satisfy this are $(1, 4)$ and $(2, 2)$. Solving these:
- For $(1, 4)$:
$$k - c = 1$$
$$k + c = 4$$
Solving these, we get:
$$k = \frac{1 + 4}{2} = 2.5$$
$$c = \frac{4 - 1}{2} = 1.5$$
This does not yield integer solutions.
- For $(2, 2)$:
$$k - c = 2$$
$$k + c = 2$$
Solving these, we get:
$$k = \frac{2 + 2}{2} = 2$$
$$c = \frac{2 - 2}{2} = 0$$
Substituting $c = 0$ back into the quadratic equation:
$$a^2 = 1$$
$$a = \pm 1$$
Thus, $a = b = \pm 1$ and $c = 0$.

- **Case 2: $a + b + c = 0$**
$$c = -(a + b)$$
Substitute $c = -(a + b)$ into the original equations:
$$a^2 = b(-a - b) + 1$$
$$a^2 = -ab - b^2 + 1$$
$$a^2 + ab + b^2 = 1$$
This is a quadratic Diophantine equation. We need to find integer solutions for $a$ and $b$ such that:
$$a^2 + ab + b^2 = 1$$
Testing small integer values:
- For $a = 1$:
$$1 + b + b^2 = 1$$
$$b^2 + b = 0$$
$$b(b + 1) = 0$$
$$b = 0 \text{ or } b = -1$$
Thus, $(a, b) = (1, 0)$ or $(1, -1)$.
- For $a = -1$:
$$1 - b + b^2 = 1$$
$$b^2 - b = 0$$
$$b(b - 1) = 0$$
$$b = 0 \text{ or } b = 1$$
Thus, $(a, b) = (-1, 0)$ or $(-1, 1)$.

Combining these results, we have the solutions:
$$(a, b, c) = (1, 0, -1), (1, -1, 0), (-1, 0, 1), (-1, 1, 0)$$

The final answer is $\boxed{ (a, b, c) = (1, 0, -1), (1, -1, 0), (-1, 0, 1), (-1, 1, 0) }$