Given a bijection $f:\mathbb Z \to \mathbb Z$ where $\mathbb Z$ is the set of all integers, does there always exist two bijections $g:\mathbb Z \to \mathbb Z$ and $h:\mathbb Z \to \mathbb Z$ which satisfy $f(n) = g(n) + h(n), ~ \forall n \in \mathbb Z$ ?
Evidently it does while $\mathbb Z$ is replaced with the set of all rational numbers $\mathbb Q$, since $f(q) = 2f(q) - f(q), ~ \forall q \in \mathbb Q$. Nevertheless, $2f$ doesn't maps $\mathbb Z$ onto $\mathbb Z$, thus it's not a bijection on $\mathbb Z$; therefor we need a new construction for the $\mathbb Z$ case.
Firstly we note that $0 = 0 + 0$ and \begin{align*} 1 &= (-2) + 3, &\qquad -1 &= 2 + (-3),\\ 2 &= 3 + (- 1), &-2 &= (-3) + 1,\\ 3 &= 1 + 2, &-3 &= (-1) + (-2). \end{align*}
It's easy to show that any bijections on the symmetric integer interval $[-3,3]$ can be represented as a sum of two bijections; or equivalently, all integers $a \in [-3,3]$ can be represented as a sum of two others, i.e., $a = b + c$ where $b, c \in [-3, 3]$, while $a,b,c$ exhausted the integer interval $[-3,3]$ respectively, and keeps $0 = 0 + 0$. For convenience, let's denote this fact as $[-3,3] = [-3,3] + [-3,3]$.
Suppose now we have $[-m, m] = [-m, m] + [-m, m]$, then we may obtain $$[-m + aL, ~ m + aL] = [-m + bL, ~ m + bL] + [-m + cL, ~ m + cL],\qquad L = 2m+1,$$ where $a,b,c \in [-n,n]$ for some positive integers $n$, provided $a = b + c$. Well, if $a,b,c$ form a representation, say, $[-n,n] = [-n,n] + [-n,n]$, then we may get a "larger" representation $$[-m - nL, ~ m + nL] = [-m - nL, ~ m + nL] + [-m - nL, ~ m + nL], \qquad L = 2m+1,$$ through the repeatedly use of "small" representations; and obviously, the larger representation "contains" the small representation as a "subrepresentation", since $0 = 0 + 0$. From now on, we may construct a representation $\mathbb Z = \mathbb Z + \mathbb Z$ by induction, start from $[-3,3] = [-3,3] + [-3,3]$. Thus we've proved that any bijections on $\mathbb Z$ can be represented as a sum of two bijections on $\mathbb Z$.
Now a problem rises:
for any integers $n > 2$, is there always a representation $[-n, n] = [-n, n] + [-n, n]$ ?
Note that by our means of "representation", $0 = 0 + 0$ is required; otherwise it's trivial; note also the above approach we've used for resolving the "$\mathbb{Z}$-representation problem" can't reach all cases.
Anyone can answer this question? Please help me.