All are Integers

We can take two cases:

Case 1: When $a=b$ .

Then we get the equation $a^2=ac+4$

$\Longrightarrow a(a-c)=4$

It makes

$\Rightarrow a=1, a-c=4$

$\Rightarrow a=-1, a-c=-4$

$\Rightarrow a=4, a-c=1$

$\Rightarrow a=-4, a-c=-1$

$\Rightarrow a=2, a-c=2$

$\Rightarrow a=-2, a-c=-2$

Thus we get $(a, b, c) = (1, 1, -3),(-1, -1, 3), (4, 4, 3), (-4, -4, -3); (2, 2, 0), (-2, -2, 0).$

Case 2:When $a\neq b$

Subtracting the first relation from the second we get

$\Rightarrow a^2 - b^2 = c(b - a).$

This gives $a+b=-c$

Substituting this in the first equation, we get

$\Rightarrow a^2 = b(-a - b) + 4.$

Thus $a^2 + b^2 + ab = 4$

Multiplication by 2 gives $(a + b)^2 + a^2 + b^2 = 8$

Thus $(a, b) = (2, -2), (-2, 2), (2, 0), (-2, 0), (0, 2), (0, -2).$

We get respectively $c =0, 0, -2, 2, -2, 2.$

Thus from both the cases we get $6$ triples each.

So the answer is $\boxed{12}$

$\begin{cases} a^2 = bc + 4 & ...(1) \\ b^2 = ca + 4 & ...(2) \end{cases} \implies (1)-(2): \ a^2 - b^2 = c(b-a) \implies (a-b)(a+b) = -c(a-b)$

Case 1: If $a-b \ne 0$ or $a \ne b$ , then $(a-b)(a+b) = -c(a-b) \implies c = -(a+b)$ . Substituting in $(1)$ .

$\begin{aligned} a^2 & = -b(a+b) + 4 \\ b^2 + ab + a^2 - 4 & = 0 & \small \color{#3D99F6} \text{It is a quadratic equation of }b. \\ \implies b & = \frac {-a \pm \sqrt{16-3a^2}}2 \end{aligned}$

We note that $b$ is integral only when $16-3a^2$ is a perfect square and $|a| < 3$ . The acceptable integral solutions are:

$\begin{cases} a = - 2 & \implies \begin{cases} b = 2 & \implies c = 0 & \implies (-2, 2, 0) \\ b = 0 & \implies c = 2 & \implies (-2, 0, 2) \end{cases} \\ a = 0 & \implies \begin{cases} b = 2 & \implies c = -2 & \implies (0, 2, -2) \\ b = -2 & \implies c = 2 & \implies (0, -2, 2) \end{cases} \\ a = 2 & \implies \begin{cases} b = 0 & \implies c = -2 & \implies (2, 0,-2) \\ b = -2 & \implies c = 0 & \implies (2, -2, 0) \end{cases} \end{cases}$

Case 2: If $a-b = 0$ or $a = b$ , then $(1) = (2) = a^2 = ca + 4$ , then

$\begin{aligned} a^2 & = ca + 4 \\ a^2 - ca - 4 & = 0 & \small \color{#3D99F6} \text{It is a quadratic equation of }a. \\ \implies a & = \frac {c \pm \sqrt{c^2+16}}2 \end{aligned}$

Again $a$ is integral only when $c^2+16$ is a perfect square and from Pythagorean triples, there is no perfect square for $|c| >3$ . The acceptable integral solutions are:

$\begin{cases} c = -3 & \implies \begin{cases} a = b = 1 & \implies (1,1,-3) \\ a = b = - 4 & \implies (-4,-4,-3) \end{cases} \\ c = 0 & \implies \begin{cases} a = b = 2 & \implies (2,2,0) \\ a = b = - 2 & \implies (-2,-2,0) \end{cases} \\ c = 3 & \implies \begin{cases} a = b = 4 & \implies (4,4,3) \\ a = b = 1 & \implies (-1,-1,3) \end{cases} \end{cases}$

Therefore, there are $\boxed{12}$ ordered integer triplets $(a,b,c)$ satisfying the system of equations.

Subtract the second equation from the first one to get $a^2 - b^2 = bc - ca$ and therefore $(a+b)(a-b) = c(b-a)$ . This naturally leads us to first consider the case $a \neq b$ so that we may simplify this last equation to $a+b = -c$ . Substituting this into the original equations we get (in both cases) $b^2 = -a^2 - ab + 4$ . I will write this as $a^2 + ab + (b^2 - 4) = 0$ and consider this to be a quadratic polynomial with variable $a$ . Computing the discriminant we get $16 - 3b^2$ and we can immediately get a range of values for $b$ given that for $a$ to be real, this discriminant needs to be non-negative so $b$ can only take values from $\{-2,-1,0,1,-2 \}$ . Finally, given that for $a$ to be an integer this discriminant has to be a perfect square we can, via checking, figure out that actually $b$ can only take values from $\{-2,0,2 \}$ . This means we have to solve 3 polynomial equations, but these turn out to be relatively simple:

$a^2 - 2a = 0 \implies a(a-2) = 0 \implies$ 2 solutions. $a^2 - 4 = 0 \implies a^2 = 4 \implies$ 2 solutions. $a^2 + 2a = 0 \implies a(a+2) = 0 \implies$ 2 solutions.

So we obtained 6 solutions in total. But these solutions occur under the assumption that $a \neq b$ so to be careful one should also check what happens if $a=b$ . If that is the case we get $a^2 = ac + 4 \implies a = c + \frac{4}{a} \implies c = a - \frac{4}{a}$ . If we now consider $c$ to be a function of $a$ it is clear that for $c$ to be an integer, $a$ must be a divisor of 4, of which there are 6 over the integers ( $-4,-2,-1,1,2,4$ ) so we obtain a new set of 6 solutions. Adding up to 12 solutions.