Alexheinis
#old riddle solution
smoky fossilBOT
smoky heron
We assume $n\ge 2$ and write $c_1,\cdots,c_n$ for the colours. Also $V_k$ is the set of numbers with colour $c_k$.
Let's first consider the colouring of $Z$. We have $n$ colours hence for large $N$ we have some 3-term AP inside $[1,N]$ with the same colour. This is van der Waerden's or Roth's theorem. Let's assume this colour is $c_1$, then $f(c_1,c_1)=c_1$.
Now let's consider the case where we only colour the set of dyadic rationals $D$. Note that if $x,y\in V_1$ and $x<y$ then all numbers in $[x,y]\cap D$ have colour $c_1$. It follows that $V_1$ is bounded from below or bounded from above. Hence we can choose arbitrarily large intervals of integers within $Z\setminus V_1$. If follows that there is some colour $c_2\not=c_1$ with $f(c_2,c_2)=c_2$. Note that $V_2$ is also an interval within $D$.
Now $V_1,V_2$ cannot be unbounded in opposite directions. Indeed, suppose $\inf(V_1)=-\infty, \sup(V_2)=\infty$ and take any $x\in D$. We have $(-\infty,l]\subset V_1$ for some $l$ and $[r,\infty)\subset V_2$ for some $r$. Hence we can write $x={{p+q}\over 2}$ where $p\in V_1,q\in V_2$. Hence $x$ has colour $f(c_1,c_2)$. That is impossible since not all $x$ have the same colour. Also $V_1,V_2$ cannot be unbounded in the same direction. Hence at most one of the sets $V_1,V_2$ is unbounded and we can find arbitrarily long intervals within $Z\setminus (V_1\cup V_2)$.
Repeating the argument we find that $f(c,c)=c$ for every colour $c$.
At least one set, say $V_\alpha$, is unbounded to the left. Also at least one set, say $V_\beta$, is unbounded to the right. Now we find the same contradiction as before, namely that each $x\in D$ has colour $f(\alpha,\beta)$.
Hence we have a contradiction and no $f$ exists.
smoky fossilBOT
Alexheinis
smoky heron
<@1275298616684187749> next time ping us so we can check it :p
I will not. I just created an account and joined the server just to share the solution. My friend shared the problem with me. Now, I will be offline (maybe I'll check the riddles sometimes).
winter sundial
We assume $n\ge 2$ and write $c_1,\cdots,c_n$ for the colours. Also $V_k$ is the...
i appreciate the contribution of van der waerden's theorem to prove that f(c,c) = c for at least one c, that is interesting
however, i have to admit most of the rest of this is incorrect
ill list the reasons i guess
firstly, if x,y in V1, then not necessarily all numbers in [x,y] n D are coloured c1
for example, we can colour 0,3 red
then only the dyadic rationals in [0,3] that are multiples of 3 would be red
this is actually more incorrect if we use the original statement, so ill approve of just colouring D
(surprisingly, we get a function f for n = 2 otherwise)
winter sundial
firstly, if x,y in V1, then not necessarily all numbers in [x,y] n D are coloure...
and yeah without this the rest of it falls apart because we can have two things unbounded in the same direction