Determine all $f:\Bbb Z \to \Bbb Z$ st $f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$. (Vietnamese TST 2005) Determine all $f:\Bbb Z \to \Bbb Z$ st $$f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$$ for all $x,y,z\in\Bbb Z$.
Source : Vietnamese TST 2005
I'm sure that the only solutions are the zero function, the identity, and the negative of the identity.  First, when $x=y=z=0$ we have $f(0)=3f(0)^3$ so $f(0)=0$ is the only integer solution.  Next if $y=-x$ and $z=0$ then $f(-x)=-f(x)$ so $f$ is odd.  Then $x=y=1$ and $z=-1$ give $f(1)^3=f(1)$ so $f(1)=0,1,-1$.
Since $f$ is a soln iff $-f$ is, we can take $f(1)=0,1$.  Then $f(2)=f(1^3+1^3+0^3)=0,2$.  Then $f(3)=f(1^3+1^3+1^3)=0,3$.  Also $f(4^3-3^3-3^3)=f(10)=f(1^3+1^3+2^3)$, so $f(4)=0,4$.  Then $f(5^3-4^3-4^3)=f(-3)$ so $f(5)=0,5$.  Then $f(6)=f(2^3-1^3-1^3)=0,6$.  Then, $f(7)=f(2^3-1^3+0^3)=0,7$.  Then $f(8)=f(2^3+0^3+0^3)=0,8$.  Then $f(9)=f(2^3+1^3+0^3)=0,9$ and $f(10)=f(2^3+1^3+1^3)=0,10$.  Etc...
There should be a smart way to make induction, but I haven't found that.  May be a nice identity.  This is the closest to the claim that $n^3$ is the sum of five cubes of integers with absolute values $<|n|$:
$$n^3=6^3+(n-2)^3+(n-3)^3-(n-6)^3-(n-11)^3+(n-12)^3.$$
At least the identity above gives me $f(11)=0,11$ and $f(12)=0,12$.
 A: As requested I'm posting the above comment as an answer here.
In order to show that every $n\geq 8$ can be written as sum of five cubes with absolute value less than $n$:

*

*we can write for odd numbers $n=2k+1$ with $k\geq 4$
$$(2k+1)^3 = (2k-1)^3 + (k+4)^3 - (k-4)^3 - 5^3 - 1^3$$

*for even numbers $n\geq 8$, we can write $n=2^a m$ where $m$ is $4$, $6$, or an odd number $\geq 5$. Then $m$ can be decomposed as $m=m_1^3+m_2^3+m_3^3+m_4^3+m_5^3$ so that
$$(2^a m)^3 = (2^a m_1)^3 + (2^a m_2)^3 + (2^a m_3)^3 + (2^a m_4)^3 + (2^a m_5)^3$$
A: Using hgmath's comment, here is a way to write $n^3$ as a sum of five cubes of integers with absolute values $<|n|$ for $n=2k$ and $k\ge 8$.  We write
$$(2k)^3=(2k-4)^3+(k+7)^3-(k-9)^3-10^3-2^3.$$
For this to work, we need to have $f(0)$, $f(1)$, $f(2)$, ..., $f(15)$.  I already got $f(0)$ until $f(12)$.
For $f(13)$ and $f(15)$ we use hgmath's comment (note that hgmath's comment is good for $k\geq 4$, but I already found until $f(12)$).  For $f(14)$, we can use
$14^3=12^3+10^3+2^3+2^3$.
