Bijection = bijection + bijection on symmetric integer intervals 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.
 A: Observation:  It is sufficient to resolve the question for $f(n)=n$.
Proof idea:  If $n=g(n)+h(n)$ for all $n$, then $f(n)=g(f(n))+h(f(n))$.  Conversely, if $f(n)=g(n)+h(n)$ then $n=f(f^{-1}(n))=g(f^{-1}(n))+h(f^{-1}(n))$.
So, if some $f$ can be decomposed into two bijections $h$ and $g$, then every $f$ can be decomposed.
So resolving this question amounts to finding a permutation $g$ of $\mathbb{Z}$ for which $n \mapsto g(n)-n$ is also a permutation.  Such permutations are called orthomorphisms.
UPDATE: I asked both Ian Wanless and Tony Evans about this, and they simultaneously pointed out that the existence of such an orthomorphism was proved in:
MR0040293 (12,670f) 
Bateman, P. T.
A remark on infinite groups. 
Amer. Math. Monthly 57, (1950). 623–624. 
20.0X 
So we can conclude that, yes, it's always possible.
A: Firstly, consider the case when we are not obliged to have the equation 0=0+0.
The question becomes: given $n \geq 1$ does there exist two permutations $g,h$ of $A:=\{-n,-n+1,\ldots,n\}$ such that $n=g(i)+h(i)$ where addition is addition over the integers (and therefore is not closed on $A$).
We can interpret an instance of this question as a set of $n$ non-attacking semiqueens on an $(2n+1) \times (2n+1)$ chessboard (with rows and columns indexed by $A$) such that no semiqueen is in cell $(i,j)$ whenever $i+j \not\in A$.  [semiqueens move up-and-down, left-and-right and along one diagonal]
So the instance you cite above is illustrated by this figure:

Just as with orthomorphisms of $\mathbb{Z}_n$ we can find linear constructions:

So we can take the equations


*

*$0=n+(-n)$, $1=(n-1)+(-n+2)$, up to $n=0+n$, along with

*$-1=-n+(n-1)$, $-2=(-n+1)+(n-3)$, up to $-n=-1+(-n+1)$.


So, g(n) takes the values $\{n,n-1,\ldots,0\}$ in the first step and $\{-n,-n+1,\ldots,-1\}$ in the second and h(n) takes the values $\{-n,-n+2,\ldots,n\}$ in the first step and $\{n-1,n-3,\ldots,-n+1\}$ in the second step.  Hence both g and h are bijections of $A$, and we can clearly see that $i \mapsto g(i)-h(i)$ is also a bijection.
So, yes, I guess it is easy in this case.
Warning: Although the motivation for this question was to prove the existence of an orthomorphism of $\mathbb{Z}$, this is insufficient to prove it.

Now lets consider the case when the equation 0=0+0 is required.  We can generate some small random examples, for example, in GAP.
RandomPermutationList:=function(n)
  local A,i,j,temp;
  A:=List([1..n],k->k);
  i:=n;
  while(i>=2) do
    j:=Random([1..i]);
    if(j<i) then
      temp:=A[i];
      A[i]:=A[j];
      A[j]:=temp;
    fi;
    i:=i-1;
  od;
  return A;
end;;

TripleJQuestion2:=function(n)
  local A,B,p,q;
  A:=Concatenation([-n..-1],[1..n]);
  B:=List([1..2*n+1],i->-n-1+i);
  while(true) do
    q:=RandomPermutationList(2*n);
    p:=Concatenation(List([1..n],i->A[q[i]]),[0],List([1..n],i->A[q[n+i]]));
    if(Set(List([1..2*n+1],i->p[i]-B[i]))=B and p[n+1]=0) then return p; fi;
  od;
  return fail;
end;;

This was successful for $4 \leq n \leq 8$:





So, combined with the fact that there are many orthomorphisms of $\mathbb{Z}_n$, it seems very likely that an instance of the desired construction exists for all $n \geq 3$.
But, continuing further (i.e. finding a constructions for all $n \geq 3$) could require considerable effort.  If you're really keen on resolving this problem, you could generate lots of random examples, and hope you could generalise enough of them to cover all $n \geq 3$.  But why would you want to do this?  It's not likely to give a surprising result, and it seems to be a fairly contrived problem (yes, the problem is harder if we assume the equation 0=0+0 is required, but why would anyone be interested in this problem?).  And also, unless you somehow build these constructions recursively, you're probably not going to solve the orthomorphisms of $\mathbb{Z}$ existence question.
A: This question is ancient (for MSE), but I feel compelled to point out that an orthomorphism for $\mathbb Z$ is not even this complicated to construct; in fact it is amenable to a just-do-it proof:
As Douglas observed we can assume $f(n)=n$ without loss of generality. Now construct $g$ and $h$ in stages. In each stage we have already selected values of $g(n)$ and $h(n)$ at finitely many $n$. For each $k\in \mathbb Z$, according to your favorite enumeration of $\mathbb Z$, do:


*

*If $g(k)$ and $h(k)$ are not yet defined, then define $g(k)=i$ and $h(k)=k-i$, where $i$ is prettiest integer such that $g$ does not already hit $i$ and $h$ does not already hit $k-i$.
(This assumes that prettiness well-orders $\mathbb Z$, but since that is the only demand on the concept, the reader will be able to select a suitable aesthetic here).

*If $g$ does not yet hit $k$, then define $g(i)=k$ and $h(i)=i-k$ where $i$ is the prettiest integer such that $g(i)$ and $h(i)$ are still undefined and $h$ does not already hit $i-k$.

*As 2, but mutatis mutandis with $g$ and $h$ swapped.
After $\omega$ many steps, $g$ and $h$ are total (because of step 1), surjective (because of step 2 and 3, respectively), and injective (by construction because only values that are not already hit will be chosen).
Before iterating over the $k$s we can select finitely many (consistent) values for $g$ and $h$ just as we please -- so achieving $0=0+0$ if one wants that is trivial.
