The number of rational points is countable, and so is the number of line segments with rational endpoints. Let us list these segments as $s_1, s_2, \dots$ First pick $n$ arbitrary rational points on $s_1$ and give them distinct colors. Then do the same for $s_2$, except that we need to avoid choosing points that have already been colored. But that is possible because $s_2$ contains infinitely many rational points. Keep doing this for each $s_k$, and the desired conclusion follows. To be complete, give the uncolored points any color at the end.

Heres a more constructive proof: For any rational $\frac{a}{b}$ in its reduced form, color it $i$ if $\max\{ v_p(b) \} \equiv i \pmod n$ where the max is taken over all primes. For any segment, let the slope be $\frac{x}{y}$ and the starting point be $\frac{a}{b}$. Then consider $\frac{a}{b} + \frac{1}{q^k} \frac{x}{y} = \frac{ayq^k + bx}{q^ky}$ where $q$ is some large prime. If $q$ is $> bx,y$ then it can't divide the numerator and hence in its lowest form its $v_q$ is just $k$. Simply let $k > y$ and vary it over the residues mod $n$ and we are done.

Color the origin arbitrarily, and color a rational point $P = (x, y) \ne (0, 0)$ with color $\min\{v_2(x), v_2(y)\} \pmod{n}.$ Now, given two distinct rational points $P = \left(\frac{x_1}{z_1}, \frac{y_1}{z_1}\right)$ and $Q = \left(\frac{x_2}{z_2}, \frac{y_2}{z_2}\right)$ we will find a point $R \in \overline{PQ}$ of color $j$ for an arbitrary $1 \le j \le n.$ As $P, Q$ are distinct, one of $x_1z_2 - x_2z_1$ and $y_1z_2 - y_2z_1$ is nonzero. Hence, we can suppose WLOG that $v_2(x_1z_2 - x_2z_1) \le v_2(y_1z_2 - y_2z_1).$ Set $\lambda = v_2(x_1z_2 - x_2z_1)$ and $\mu = v_2(z_1z_2).$ Choose $k > \lambda$ so that $\lambda - \mu - k \equiv j \pmod{n}.$ Set $\alpha = 2^{-k}$ and consider $R = \alpha P + (1 - \alpha)Q \in \overline{PQ}.$ A straightforward calculation shows that \begin{align*} R = \left(\frac{2^kx_2z_1 + (x_1z_2 - x_2z_1)}{2^kz_1z_2}, \frac{2^ky_2z_1 + (y_1z_2 - y_2z_1)}{2^kz_1z_2}\right). \end{align*}Since $k > \lambda$, the $2$-adic valuation of the numerator of the $x$-coordinate of $R$ is $\lambda.$ Hence, the $2$-adic valuation of the $x$-coordinate of $R$ is $\lambda - \mu - k.$ From the assumption $v_2(x_1z_2 - x_2z_1) \le v_2(y_1z_2 - y_2z_1)$, it's easy to see that the $2$-adic valuation of the $y$-coordinate of $R$ is at least $\lambda - \mu - k.$ Hence, as $\lambda - \mu - k \equiv j \pmod{n}$, the point $R$ is of color $j$, as desired.

We consider only the $y$ coordinate. If $y=c/d$, where $gcd(c,d)=1$, then color it $cd \pmod n$. Consider any pair of rational points. There exists a prime $p$ such that there are $n$ consecutive integers, none of which are divisible by $p$, which, when divided by $p$, lie between the y-coordinate of the rational points. Consider any of those points: $i/p$. Since $i$ is not divisible by $p$, it is coloured $ip$, and all of these are distinct $\pmod n$ since $ip \equiv jp \pmod n$ is equivalent to $i \equiv j \pmod n$ i.e. $i=j$, so they hit every residue I.e. Every colour.

fattypiggy123 wrote: Heres a more constructive proof: For any rational $\frac{a}{b}$ in its reduced form, color it $i$ if $\max\{ v_p(b) \} \equiv i \pmod n$ where the max is taken over all primes. For any segment, let the slope be $\frac{x}{y}$ and the starting point be $\frac{a}{b}$. Then consider $\frac{a}{b} + \frac{1}{q^k} \frac{x}{y} = \frac{ayq^k + bx}{q^ky}$ where $q$ is some large prime. If $q$ is $> bx,y$ then it can't divide the numerator and hence in its lowest form its $v_q$ is just $k$. Simply let $k > y$ and vary it over the residues mod $n$ and we are done. But what about the slope is 0？

MathPanda1 wrote: We consider only the $y$ coordinate. If $y=c/d$, where $gcd(c,d)=1$, then color it $cd \pmod n$. Consider any pair of rational points. There exists a prime $p$ such that there are $n$ consecutive integers, none of which are divisible by $p$, which, when divided by $p$, lie between the y-coordinate of the rational points. Consider any of those points: $i/p$. Since $i$ is not divisible by $p$, it is coloured $ip$, and all of these are distinct $\pmod n$ since $ip \equiv jp \pmod n$ is equivalent to $i \equiv j \pmod n$ i.e. $i=j$, so they hit every residue I.e. Every colour. Your line $y=q\in \mathbb{Q}$ must contain only one-color-points which is incorrect

I saw this problem at the time it appeared here. Didn't post a solution since it was the same as in #2. In my opinion, this is the best approach, since it doesn't use any specific thing about the rationals nature, other than they are countably many and on any segment with rational ends there are infinitely many rationals. All the other proofs would use some NT properties of rationals and thus it makes them "weaker". Even further, this proof uses only the fact that the set of all rational points and the set of the rational points on any segment have equal cardinalities. It may not be countable. The proof is the same, only we arrange the corresponding objects (e.g. the rational points) up to some ordinal (the first one with that cardinality) and apply the same procedure, using the transfinite induction.

Here is my construction: Let $\nu_2(\frac xy)=\nu_2(|x|)-\nu_2(|y|)$ if $x,y\in \mathbb{Z}_{\ne 0}$. We let $(x,0), (0,x)$ be colored with $R(\nu_2(x),n)$ and $(p,q)$ for $p,q\in\mathbb{Q}_{\ne 0}$ be colored with $R(\max\{\nu_2(p),\nu_2(q)\}, n)$. Here, $R(a, n)$ denotes the $0\le r<n$ such that $n\mid a-r$. If the line is horizontal or vertical, it suffices to show that for fixed $\frac ab, \epsilon, e$ ($e$ is very large), I can take $c,d$ such that $|\frac ab - \frac{c2^e}{d}| < \epsilon$. To do so, we note $\frac ab - \frac{c2^e}{d}=\frac{ad-2^ebc}{db}$. Notice there exist infinitely many solutions $(c,d)$ such that $|ad-(2^eb)c|\le 2^e$ as $\gcd(a,b)=1$. Then taking one $d>2^{e^{e^e}}$ would suffice (recall $e>>n$). It suffices to show that for any line $y=px+q, p,q\in \mathbb{Q}$, there exists $l$ such that the set of points $(x,y)$ on the line with $\max\{\nu_2(x),\nu_2(y)\}=c$ for all $c>l$ is dense. We can simply take $l>\max\{\nu_2(\frac qp), \nu_2(q)\}$, then $\nu_2(px+q)=\nu_2(q)<\nu_2(x)$ so we are good.

China Team Selection Test 2016 Test 3 Day 1 wrote: In the coordinate plane the points with both coordinates being rational numbers are called rational points. For any positive integer $n$, is there a way to use $n$ colours to colour all rational points, every point is coloured one colour, such that any line segment with both endpoints being rational points contains the rational points of every colour? The answer is yes. Let the origin $O$ have the colour $0$ and for any point $P \neq O$ is a rational point with coordinates $(x,y)$ give the colour whose number is a remainder of $\min(\nu_2(x),\nu_2(y))$ when divided by $n$ Consider a line segment $\mathbf{I}$ whose endpoints are $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ and we'll prove the existence of an point on $\mathbf{I}$ whose colour is $i.$ Pick $k$ large enough to ensure $\nu_2(x_1-x_2)-k<\nu_2(x_2)$ and $\nu_2(y_1-y_2)-k <\nu_2(y_2)$ whenever $x_1 \neq x_2,y_1 \neq y_2$ Choose $Q_k=(\frac{x_1-x_2}{2^k}+x_2,\frac{y_1-y_2}{2^k}+y_2)=(x_k,y_k)$ which lies on $\mathbf{I}$ Observe that $\nu_2(x_k)=\nu_2(x_1-x_2)-k$ and similarly for $v_k$ Thus $\min(\nu_2(u_k),\nu_2(v_k))=\nu_2(u_k)=\nu_2(x_1-x_2)-k$ which covers all residues $\pmod n$,so we are done.

Sprites wrote: China Team Selection Test 2016 Test 3 Day 1 wrote: In the coordinate plane the points with both coordinates being rational numbers are called rational points. For any positive integer $n$, is there a way to use $n$ colours to colour all rational points, every point is coloured one colour, such that any line segment with both endpoints being rational points contains the rational points of every colour? The answer is yes. Let the origin $O$ have the colour $0$ and for any point $P \neq O$ is a rational point with coordinates $(x,y)$ give the colour whose number is a remainder of $\min(\nu_2(x),\nu_2(y))$ when divided by $n$ Consider a line segment $\mathbf{I}$ whose endpoints are $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ and we'll prove the existence of an point on $\mathbf{I}$ whose colour is $i.$ Pick $k$ large enough to ensure $\nu_2(x_1-x_2)-k<\nu_2(x_2)$ and $\nu_2(y_1-y_2)-k <\nu_2(y_2)$ whenever $x_1 \neq x_2,y_1 \neq y_2$ Choose $Q_k=(\frac{x_1-x_2}{2^k}+x_2,\frac{y_1-y_2}{2^k}+y_2)=(x_k,y_k)$ which lies on $\mathbf{I}$ Observe that $\nu_2(x_k)=\nu_2(x_1-x_2)-k$ and similarly for $v_k$ Thus $\min(\nu_2(u_k),\nu_2(v_k))=\nu_2(u_k)=\nu_2(x_1-x_2)-k$ which covers all residues $\pmod n$,so we are done. How can you come up with such an idea!

Transfinite induction simulator. We show this is possible for all $n$. There are at most ${\mathbb Q}^2$ lines which is countable. Enumerate the lines $\ell_1, \ell_2, \dots$, and take the natural colorings $T_1 \subseteq T_2 \subseteq T_3$ such that in $T_n$ has all of $\ell_1, \ell_2, \dots$ having $n$ colors. Then take $\bigcup_{n \ge 1} T_n$ to finish.

YaoAOPS wrote: Transfinite induction simulator. We show this is possible for all $n$. There are at most ${\mathbb Q}^2$ lines which is countable. Enumerate the lines $\ell_1, \ell_2, \dots$, and take the natural colorings $T_1 \subseteq T_2 \subseteq T_3$ such that in $T_n$ has all of $\ell_1, \ell_2, \dots$ having $n$ colors. Then take $\bigcup_{n \ge 1} T_n$ to finish. Can you explain it in detail? Thanks!

See https://blog.evanchen.cc/2015/04/10/cauchys-functional-equation-and-zorns-lemma/ and https://en.wikipedia.org/wiki/Transfinite_induction for references on transfinite induction. But the idea is since we can keep on construct a consistent coloring for any finite number of lines, taking the coloring we get as that goes on to "infinity" then works.