Irish Mathematical Olympiad

There are infinitely many solutions. We first examine some constraints.
If $f(n) = f(m)$ for $n, m \ge 0$, then $4n^2 = 4m^2$ and so $n = m$. Similarly $n = m$ if $f(n) = f(m)$ for $n, m \le 0$. Now let $a = f(0)$ and substitute $n = 0$ in the given functional equation. We obtain $f(a+1) = f(a-1)$, hence $a+1$ and $a-1$ must have different signs. This forces $f(0) = a = 0$.
Requiring the given functional equation for all $n \in \mathbb{N}$ (positive integers) makes sense as a condition on functions $f: \mathbb{N} \to \mathbb{N} \setminus \{1\}$. If we can find such a function $f: \mathbb{N} \to \mathbb{N} \setminus \{1\}$, then we get a solution to our original problem when we extend it to a function defined on $\mathbb{Z}$ by letting $f(0) = 0$ and $f(-n) = f(n)$, $n \in \mathbb{N}$.
We already know that $f$ must be injective (on $\mathbb{N}$), so one possibility is to take $f(n)$ to be a distinct multiple of 4 for distinct $n \in \mathbb{N}$. Then the given functional equation just gives a sequence of uncoupled restrictions on $f$. The idea is therefore to pick some simple function $g: \mathbb{N} \to 4\mathbb{N}$ and write $f(n) = g(n)$ for most values of $n \in \mathbb{N}$, but instead use the given equation.
$$f(f(m) - 1) = f(f(m) + 1) - 4m^2$$
to define $f(n)$ for some numbers $n \equiv 3 \pmod 4$. Defining $f(n)$ this way for $n = f(m) - 1$ in terms of $f(n+2) = g(n+2)$ certainly gives a value that is still a multiple of 4. Any monotonic function $g$ that increases quickly enough to ensure that $f(f(m) + 1) - 4m^2 > f(f(m))$ will suffice. For example, fix an integer $K > 0$ and take $g(n) = 4Kn^2$, and so, for $n > 0$ we define
$$f(n) = \begin{cases} 4K(n+2)^2 - 4m^2, & \text{if } n \equiv 3 \pmod 4 \text{ and } n = f(m) - 1 \\ & \text{for some } m \in \mathbb{N}, m < n; \\ 4Kn^2, & \text{otherwise.} \end{cases}$$
This equation gives $f(1) = 4K$, and also defines $f(n)$, $n > 1$, in terms of $f(m)$ for $m \in \mathbb{N}$, $m < n$. Thus it assigns exactly one value $f(n)$ to $n \in \mathbb{N}$. Inductively we see that $f(n) \ge 4Kn^2$ for all $n \in \mathbb{N}$. This inequality implies that $f(m) \ge 4Km^2 \ge 4m^2$ and so, in the exceptional case $n = f(m) - 1$,
$$f(n) = 4K(n + 2)^2 - 4m^2 \ge 4K(n + 2)^2 - n - 1 > 4(n + 1)^2 = f(n + 1)$$
and obviously $f(n) < 4K(n + 2)^2 = f(n + 2)$. Thus $f$ is injective. Because $f(m) - 1 > m$ for $m > 0$ it is also now clear that $f$ satisfies the given equation.