I think the following works. It's easy to see that $k\geq3$ (Consider values $0,5,12$). Now our function will be: for $x \equiv 0,1,2,3,4 \mod 15$ put $f(x)=1$ for $x \equiv 5,6,7,8,9 \mod 15$ put $f(x)=2$ for $x \equiv 10,11,12,13,14 \mod 15$ put $f(x)=3$ It works

Megus wrote: I think the following works. It's easy to see that $k\geq3$ (Consider values $0,5,12$). Now our function will be: for $x \equiv 0,1,2,3,4 \mod 15$ put $f(x)=1$ for $x \equiv 5,6,7,8,9 \mod 15$ put $f(x)=2$ for $x \equiv 10,11,12,13,14 \mod 15$ put $f(x)=3$ It works I don't think so... $f(21) = f(9)$ and $21-9 = 12$. EDIT: I think if you change it to $\mod 17$ you can get $k = 4$, which I think is the minimum.

k=4 and $f(x)=1,x\equiv1,3\pmod{8}$,$f(x)=2,x\equiv2,4\pmod{8}$,$f(x)=3,x\equiv5,7\pmod{8}$,$f(x)=4,x\equiv6,0\pmod{8}$.

$ k=1 $ is not a solution. $ k=2 $ is not a solution because if a is red,a+7 is blue then a+12=(a+7)+5 can't be red or blue false. $ k=3 $ isn't a solution because if a is red,a+5 blue,a+12 green then a+7 can't be green or red so is blue. so a+5 and a+7 are of the same colour so the even numbers are of a colour and the odd of the other false. $ k=4 $ is the solution, the even numbers take 2 numbers and the odd other 2. the numbers congruent with 0,2,4,6,8,10 mod 24 take a value and the numbers congruent with 12,14,16,18,20,22 take the other value. for the odd numbers we take a value for the numbers congruent with 1,3,5,7,11 and the other for the numbers congruent with 13,15,17,19,21,23 mod 24.