5. Each rational number is painted with one of two colors, white and red. Such a coloring is called sanferminera when for any two rational numbers $x, y$, with $x \neq y$, if one of the following three conditions is met:
a) $x y=1$,
b) $x+y=0$,
c) $x+y=1$,

then $x$ and $y$ are painted in different colors. How many sanferminera colorings are there?

## SOLUTION:

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
If a coloring is sanferminera, we can find another sanferminera coloring by simultaneously swapping the color of each rational, from red to white and from white to red; if in the initial coloring two rationals have different colors, they will also have different colors in the resulting one. Let's then find the number of sanferminera colorings such that, without loss of generality, the rational 1 is painted red, and the total number of sanferminera colorings will be exactly double.

Suppose then that we have a sanferminera coloring with 1 painted red. Given the color of the rational $x>0$, the color of the rational $-xu^{\prime}$, and after at most $u$ steps, we would reach $u^{\prime}=0$, that is, to an integer, which does have an associated color. Note also that the process is unique since, at each step, the values of $m, u, v$ are uniquely determined by the value of the positive rational $q$, so with all positive integers being red, each positive rational will have one and only one associated color. This unique coloring for positive rationals extends uniquely to all rationals, as we have already seen. There is then at most one sanferminera coloring painting the rational 1 red.

Let's check that indeed this unique coloring we have constructed satisfies the conditions of the statement:

1) If $x+y=0$ with $x \neq y$, without loss of generality $x>0>y=-x$, so $y$ has, by the way of extending the coloring to all rationals, a different color from $x$.
2) If $x y=1$ with $x \neq y$, either both are positive, or both are negative, having in the second case $x, y$ colors respectively opposite to those of $-x,-y$, which would be positive, reducing the verification to the first case. If $x, y$ are both positive, without loss of generality $x>1>y$, with $y=0+\frac{u}{v}$ for positive integers $uy$, and either $x=1$ is red, $y=0$ is white, or $x>1>0>y$, or $1>x>y>0$. In the second case, the positive rational $x-1$ has, by construction, the same color as $x$, so by 1), $x$ has the opposite color of $1-x=y$. In the third case, there exist positive integers $u<v$ coprime such that $x=\frac{v}{u+v}, y=\frac{u}{u+v}$. By construction and by 2), $x$ has the opposite color of $\frac{u+v}{v}=1+\frac{u}{v}$, so $x$ has the same color as $\frac{v}{u}$. But also by construction and by 2), $y$ has the opposite color of $y=\frac{u+v}{u}=1+\frac{v}{u}$, so opposite to $\frac{v}{u}$, and opposite to $x$.

Thus, the constructed coloring is sanferminera, and it is the only one that can be sanferminera with the rational 1 painted red. Restoring generality, there are exactly two sanferminera colorings.