, 2018

(1) $f(f(x)) = 2x + f(-x), \forall x \in \mathbb{Z}$
Note that, if $a \in \mathbb{Z}$ satisfies $f(a) = 0$, then substituting $x, y = a$ into the condition,
$$4a^2 = f(a + f(a))^2 = f(a)^2 = 0.$$
We see that $a = 0$. Now substitute $y = 0$ to get
$$f(x)f(f(x)) = f(x)(2x + f(-x)).$$
So if $x \neq 0$, the conclusion holds. On the other hand, when $x = 0$, it obviously also holds.

(2) $f(x + f(x)) = 2x, \forall x \in \mathbb{Z}$
We only need to consider the case where $x$ is not $0$. Substitute $y = x$ into the condition:
$$f(x + f(x))^2 = 4x^2 \Rightarrow f(x + f(x)) = 2x \text{ or } -2x.$$
Suppose there exists $b \in \mathbb{Z} - \{0\}$ such that $f(b + f(b)) = -2b$. Substituting $b, 2b + f(b)$ for $x, y$ in the condition, we get
$$f(b + f(2b + f(b)))f(2b + 2f(b)) = 0.$$
From the injectivity of $f$ at $0$, we know:
$$b + f(2b + f(b)) = 0 \text{ or } 2b + 2f(b) = 0,$$
but the latter cannot hold, otherwise we would have $f(b) = -b$, and thus
$$f(-b) = f(f(b)) = 2b + f(-b) \Rightarrow b = 0.$$
Hence $f(2b + f(b)) = -b$. Therefore
$$4b^2 = f(b + f(b))^2 = f(2b + f(b) + f(2b + f(b)))^2 = 4(2b + f(b))^2.$$
That is, $f(b) = -b$ or $f(b) = -3b$. From the earlier discussion we know the only possibility is $f(b) = -3b$, but then we would have
$$f(-b) = f(2b + f(b)) = -b \Rightarrow f(b) = f(f(-b)) + 2b = b.$$
So $b=0$, a contradiction. Combining the above, the sub-conclusion holds.

(3) $f$ is an injective odd function and $f(2x) = 2f(x)$, $\forall x \in \mathbb{Z}$
First, substituting $(-x)$, $f(x)$ for $x, y$ in the original problem gives
$$f(-x + f(f(x)))f(f(x) + f(-x)) = 0.$$
So $f(f(x)) = x$ or $f(x) = -f(-x)$. But even if the former holds
$$\begin{aligned} & f(-x) = f(f(x)) - 2x = -x \\ \Rightarrow & f(x) = f(f(-x)) + 2x = x = -f(-x). \end{aligned}$$
The above discussion tells us $f$ is an odd function. Then the first sub-conclusion becomes
$$f(f(x)) = 2x - f(x).$$
Thus $f$ is injective, and finally by the second sub-conclusion
$$f(2x) = f(f(x)) + f(f(x)) = 2f(x).$$

(4) If there exists an odd number in $f(\mathbb{Z})$, then $f(x) = x, \forall x \in \mathbb{Z}$
Suppose the integer $c$ satisfies that $f(c)$ is odd, then substituting $y = x + c$ into the original problem gives
$$f(x + f(x + c))f(x + c + f(x)) = (2x + f(c))(2x + 2c - f(c)).$$
In particular, $f(x + c + f(x))$ is odd. Using $f(2x) = 2f(x)$ we know:
$$x + f(x) \equiv 0 \pmod{2}.$$
Then using $f(2x) = 2f(x)$ again we can deduce that
$$v_2(f(x)) = v_2(x), \forall x \in \mathbb{Z},$$
Substituting $y = \frac{x - f(x)}{2}$ gives
$$\begin{aligned} & f\left(x + f\left(\frac{x - f(x)}{2}\right)\right) f\left(\frac{x + f(x)}{2}\right) \\ & = \left(2x + f\left(\frac{-x - f(x)}{2}\right)\right) \left(x - f(x) + f\left(\frac{x + f(x)}{2}\right)\right). \end{aligned}$$
Note that $f(\frac{x+f(x)}{2}) = x$, so
$$f\left(x + f\left(\frac{x - f(x)}{2}\right)\right) = 2x - f(x) = f(f(x)).$$
By injectivity, $f(\frac{x-f(x)}{2}) = f(x) - x$, so
$$v_2(f(x) - x) = v_2\left(f\left(\frac{x - f(x)}{2}\right)\right) = v_2\left(\frac{x - f(x)}{2}\right).$$
The only possibility is $f(x) = x$

(5) If $f(\mathbb{Z})$ contains only even numbers, then $f(x) = -2x, \forall x \in \mathbb{Z}$
We can consider a new function $g: \mathbb{Z} \to \mathbb{Z}$ satisfying $2g(x) = f(x), \forall x \in \mathbb{Z}$, then the condition can be rewritten as
$$g(x + 2g(y))g(y + 2g(x)) = (x + g(y - x))(y + g(x - y)).$$
Substituting $-f(x)$ for $y$ in the above, we get
$$g(x+2g(-g(x)))g(g(x)) = (x+g(-g(x)-x))(-g(x)+g(x+g(x))).$$
Since $g$ inherits most of the properties of $f$, we have
$$\left(\frac{x - g(x)}{2}\right)^2 \ge \text{RHS} = \text{LHS} = g(g(x))^2 = \left(\frac{x - g(x)}{2}\right)^2.$$
Thus $g(x + g(x)) = \frac{x + g(x)}{2}$. But $g(2x) = 2g(x)$ tells us
$$v_2(g(x)) \geq v_2(x), \forall x \in \mathbb{Z}.$$
So
$$g(x) = -x, \forall x \in \mathbb{Z} \Rightarrow f(x) = -2x, \forall x \in \mathbb{Z}.$$

Remark: In fact, the condition $f(0) = 0$ is not needed; it can be derived using just the original condition. But stating it directly in the problem statement can reduce some computation. Below is the proof that $f(0) = 0$: substituting $x, y = 0$ gives
$$f(f(0))^2 = f(0)^2 \Rightarrow f(f(0)) = f(0) \text{ or } -f(0).$$
If $f(f(0)) = -f(0)$, then substituting $x, y = f(0)$ gives
$$f(0)^2 = 9f(0)^2 \Rightarrow f(0) = 0.$$
If $f(f(0)) = f(0)$, then likewise substituting $x, y = f(0)$ gives
$$f(2f(0))^2 = 9f(0)^2 \Rightarrow f(2f(0)) = 3f(0) \text{ or } -3f(0).$$
(i) $f(2f(0)) = 3f(0)$: substituting $x = f(0), y = 0$ gives
$$f(2f(0))f(f(f(0))) = (2f(0) + f(-f(0)))f(f(0)).$$
If $f(0) \neq 0$ then $f(-f(0)) = f(0)$. Then substituting $x = -f(0), y = 0$:
$$f(0)^2 = f(0)(f(f(0)) - 2f(0)) \Rightarrow f(0) = 3f(0).$$
Thus $f(0) \neq 0$ is a contradiction, so $f(0) = 0$.

(ii) $f(2f(0)) = -3f(0)$: at this point, again substituting $x = f(0), y = 0$ gives
$$f(-f(0)) = -5f(0)$$
Then substituting $x = -f(0), y = 0$ gives
$$f(f(-f(0)))f(0) = -f(0)f(-f(0)).$$
Therefore when $f(0) \neq 0$, $f(-5f(0)) = 5f(0)$, and finally taking $x, y = 5f(0)$ gives
$$f(0)^2 = 81f(0)^2$$
which contradicts $f(0) \neq 0$, so in this case, $f(0) = 0$
Combining the above, $f(0) = 0$.