I vaguely remember having seen this problem somewhere else, but not vividly enough to prevent me spending a while finding the solution. We can wlog let $ f(n)=n$ (by permuting the arguments of u and v accordingly). I claim the following algorithm generates bijections u and v summing to identity, where I make both odd functions (u(-x)=-u(x) etc) and u(0)=v(0)=0. For k, starting at k=0 and continuing upwards, let $ u(2k+1)=-m, v(2k+2)=-l$ where $ m$ is the least positive integer such that $ u(x)$ does not already contain $ m$ in its image between $ -2k$ and $ 2k$, and similar for $ v,l$, and correspondingly let $ v(2k+1)=2k+1+m, u(2k+2)=2k+2+l$ Suppose this algorithm fails to generate a bijection. It clearly can exclude no integer from its image, because it specifically demands inclusion of all integers not yet in its image, so it will only break if injectivity fails. Suppose k is the least integer for which we obtain u(x)=u(y) or v(x)=v(y) for |x|,|y|<=2(k+1). If u(x)=u(y) then 2k+2+l = u(z) for |z|<2k+2. z must be odd, for obvious reasons (if it's even, l hasn't been chosen correctly). This would imply that the image of u(x) as x varies on -(|z|-1),|z|-1 must contain all the positive integers between 1 and 2k+2+l. This is clearly absurd, since |z|-1<2k+1<2k+2+l. We get similar contradictions in the v(x)=v(y) case, and thus conclude that u,v are both bijections, as required.

I think we can do even less work. Draw a table containing three rows. The first row contains the integers (in, say, the order 0,1,-1,2,-2,...), while the second and third rows shall contain the values of u(n) and v(n) respectively (n being the value in the first row). We repeat the following process to fill in all the columns: i)look at the first column which we have not yet filled, with first row n. Choose u(n) and v(n) to have absolute value larger than anything we have yet chosen so that they sum to f(n). ii)look at the first number, k, which we have not yet chosen as a value u(i). Choose n such that f(n)-k has not yet been chosen as a value for v, and fill in the n-column accordingly. iii)repeat ii with u and v reversed. Cycling through i, ii and iii generates bijections u, v with u+v=f Jack

is this trivial simply define v=f(f)+f and u=-f(f).

They are injective, but not surjective.

Here is a slightly more explicit construction: Note that we only need to consider $f$ as the identity function by a suitable permutation Now consider, $$U=[-2,1,-3,0,3,-1,2]^n$$$$V=[-1,-3,2,0,-2,3,1]^n$$(By $[\dots]^n$ we mean placing copies of $[\dots]^{n-1}$ each shifted by a corresponding entry of $[\dots]\cdot|[\dots]^{n-1}|$. For example, $$[-2,1,-3,0,3,-1,2]^2=[-16,-13,-17,-14,-11,-15,-12,5,8,4,7,10,6,9,-23,-20,-24,-21,-18,-22,-19,-2,1,-3,0,3,-1,2,19,22,18,21,24,20,23,-9,-6,-10,-7,-4,-8,-5,12,15,11,14,17,13,16]$$Centering both $U$ and $V$ at $0$ will result in two bijections that sum to the identity.

vjdjmathaddict wrote: is this trivial simply define v=f(f)+f and u=-f(f). Why is $f(f)+f$ bijective?