Is this the only solution? Write $(a_2, ..., a_n)=d, a_1=d'$ where $(d, d')=1$. Take positive reals $0=c_0<c_1<c_2<...<c_{dd'}$. $$f(x)= \left\lfloor \frac{x}{dd'} \right\rfloor c_{dd'} +c_{ \left\{ \frac{x}{dd'} \right\} dd'}$$ The hard part is to prove for $n=2$. It is obvious that $f(0)=0$, thus putting $x_1=0$ gives us $$f \left( \sum_{i=2}^{n} a_ix_i \right) = \sum_{i=2}^{n} f(a_ix_i) \Rightarrow f \left( \sum_{i=1}^{n} a_ix_i \right) = \sum_{i=1}^{n} f(a_ix_i) = f(a_1x_1)+f \left( \sum_{i=2}^{n} a_ix_i \right) $$ By Bezout, we have $f(a_1x_1+dx)=f(a_1x_1)+f(dx)$ where $(a_2, ..., a_n)=d$ and $\forall x \in \mathbb{Z}$. It is also obvious that $(a_1, d)=1$, otherwise we would have a prime dividing all $a_i$s, contradiction to the given condition. Take $a_1=d'$ and we have $f(dx+d'x')=f(dx)+f(d'x')$ for coprime $d, d'$ and all $x, x'$. Write $x=dd't+r$ and $x'=dd't'+r'$; $t, t', r, r'$ are yet to be determined. (We are not using Euclidean Division here, be careful) $$f(d^2d't+dr+dd'^2+d'r')=f(d^2d't+dr)+f(dd'^2t'+d'r')=f(d^2d't)+f(dr)+f(dd'^2t')+f(d'r')=f(dd'(dt+d't'))+f(dr+d'r')$$ Take $y=dd'm+s$ where $0 \leq s <d'$, and choose $t, t', r, r'$ such that $dt+d't'=m$ and $dr+d'r'=s$ by Bezout as $d, d'$ are coprime. We clearly have $f(dd'm+s)=f(dd'm)+f(s)$. It is obvious that if we construct $0=c_0=f(0), c_1=f(1), ..., c_{dd'}=f(dd')$ as we want, in increasing order, we can use the last result to finish for every integer that $ \left\lfloor \frac{x}{dd'} \right\rfloor c_{dd'} +c_{ x \pmod{dd'}}$

electrovector wrote: Is this the only solution? Write $(a_2, ..., a_n)=d, a_1=d'$ where $(d, d')=1$. Take positive reals $0=c_0<c_1<c_2<...<c_{dd'}$. $$f(x)= \left\lfloor \frac{x}{dd'} \right\rfloor c_{dd'} +c_{ \left\{ \frac{x}{dd'} \right\} dd'}$$ I Agree with the rest but does this categorise all the functions? There are some conditions on the $c_i$ and the $f(i)$, you define the $c_i$'s in terms of $f(i)$'s but we don't know what the $f(i)$'s really are. For example if the problem were to ask us which numbers satisfied $x^{p-1} \equiv 1 \pmod{p}$ this would be like 'check all numbers between $1$ and $p$ and the numbers satisfying are whose remainders by $p$ lie in that set'.

Bump ! Can someone post a complete solution ? Not all of the solutions mentioned by electrovector verify because to verify the condition $f(dx+d'x')=f(dx)+f(d'x')$ the $c_i$ must verify that 1) $c_{db+d'b'}$=$c_{db}+c_{d'b'}$ whenever $db+d'b'<dd'$ 2)$c_{dd'}+c_{db+d'b-dd'}$=$c_{db}+c_{d'b'}$ whenever $db+d'b'\ge dd'$ and this are the answers to the case $n=2$ but i don't if you can reduce it even more to a more accurate answer and i also don't know how to solve the general case to verify (for example when $d=2,d'=3, f(0)=0;f(1)=a;f(2)=b;f(3)=c;f(4)=d;f(5)=c+b;f(6)=c+d-a$ for every $0<a<b<c$ and $d \in (max \left\{c,a+b \right\},b+c)$ so there are "weird" solutions.

No one ? It is a good problem.

Official Solution?