Problem 1596

Given the polynomial $f(x,y,z)$ with integer coefficients, we know that for all real numbers $x, y, z$, the following equation holds:
$$f(x,y,z) = -f(x,z,y) = -f(y,x,z) = -f(z,y,x)$$

We need to prove that if $a, b, c \in \mathbb{Z}$, then $f(a,b,c)$ takes an even value.

1. Symmetry and Antisymmetry Properties:
From the given conditions, we can derive the following properties:
$$f(x,y,z) = -f(x,z,y)$$
$$f(x,y,z) = -f(y,x,z)$$
$$f(x,y,z) = -f(z,y,x)$$

2. Substitution and Simplification:
Let's substitute specific values to understand the behavior of $f$:
- Set $x = y = m$ and $z = n$:
$$f(m,m,n) = -f(m,n,m) = -f(n,m,m)$$
Since $f(m,m,n) = -f(m,n,m)$ and $f(m,m,n) = -f(n,m,m)$, we have:
$$f(m,m,n) = 0$$
- Set $m = n = 0$:
$$f(0,0,0) = 0$$

3. Generalization to Even and Odd Values:
- For $m = 0$ and $n = 1$:
$$f(0,0,1) = -f(0,1,0) = -f(1,0,0)$$
Since $f(0,0,1) = -f(0,1,0)$ and $f(0,0,1) = -f(1,0,0)$, we have:
$$f(0,0,1) = 0$$
- For $m = n = 1$:
$$f(1,1,1) = -f(1,1,1)$$
This implies:
$$2f(1,1,1) = 0$$
$$f(1,1,1) = 0$$

4. Conclusion for General Integer Values:
- For any integers $a, b, c$:
$$f(a,b,c) \equiv 0 \pmod{2}$$
This means $f(a,b,c)$ is even for all integer values of $a, b, c$.

The final answer is $\boxed{ f(a,b,c) \equiv 0 \pmod{2} }$