We are given the functional equation for $f: \mathbb{Z} \to \mathbb{Z}$:
$$2f(a^2 + b^2 + c^2) - 2f(ab + bc + ca) = f(a - b)^2 + f(b - c)^2 + f(c - a)^2$$
for any integers $a, b,$ and $c$.

To find all such functions $f$, we proceed as follows:

### Step 1: Initial Analysis
Let $P(a, b, c)$ denote the given assertion. Adding $P(a, b, c)$ and $P(-a, -b, -c)$ yields:
$$2f(a^2 + b^2 + c^2) - 2f(ab + bc + ca) = f(a - b)^2 + f(b - c)^2 + f(c - a)^2.$$
This simplifies to:
$$\sum f((a - b)^2) = \sum f(a - b)^2,$$
indicating that $f(x^2) = f(x)^2$.

### Step 2: Case Analysis
#### Case 1: $f(1) = 0$
If $f(1) = 0$, then $P(1, 0, 0)$ implies $f(-1) = 0$. By induction and using the relation $Q(x, -x-1, 1)$, we find that $f(x) = 0$ for all $x$.

#### Case 2: $f(1) \neq 0$
If $f(1) \neq 0$, then $P(1, 0, 0)$ gives $2f(1) = f(1)^2 + f(-1)^2$. This implies $f(1) = 1$ and $f(-1) = -1$. Using induction and the relations derived from $P(x, 1, 0)$ and $P(x, -1, 0)$, we find that $f(x) = x$ for all $x$.

### Conclusion
The only functions that satisfy the given functional equation are:
1. $f(x) = 0$ for all $x$.
2. $f(x) = x$ for all $x$.

The answer is: $\boxed{f(x)}$ = 0 $\text{}$ or } f(x) = x}.