25. Every positive integer is to be coloured according to the following three rules. (i) Each number is to be coloured either red or green. (ii) The sum of any two different red numbers is a red number. (iii) The sum of any two different green numbers is a green number. In how many different ways can this be done?
A 0
B 2
C 4
D 6
E more than 6
25. D There are just six ways to colour the positive integers to meet the two conditions:
(a) All positive integers are coloured red.
(b) 1 is coloured red and all the rest are coloured green.
(c) 1 is coloured red, 2 is coloured green and all the rest are coloured red.
(d) All positive integers are coloured green.
(e) 1 is coloured green and all the rest are coloured red.
(f) 1 is coloured green, 2 is coloured red and all the rest are coloured green.
Note that (d), (e) and (f) can be obtained from (a), (b) and (c) respectively by swapping round the colours red and green.
It is easy to see that these colourings meet the two conditions given. We now need to show that no other colouring does so.
We first show there is no colouring different from a), b) and c) in which the number 1 is coloured red. In such a colouring, since it is different from b), there must be some other number coloured red. Let $n$ be the smallest other number coloured red. Then, as 1 and $n$ are red, so is $n+1$. Hence $(n+1)+1=n+2$ is also red, and so on. So all the integers from $n$ onwards are red. Since such a colouring is different from a), $n \neq 2$, and since it is different from c) $n \neq 3$. So $n \geqslant 4$ and therefore 2 and 3 are green. But then $2+3=5$ is green, so
$2+5=7$ is green and so on. So all positive integers of the form $2 k+3$ are green. In particular $2 n+3$ is green, contradicting the fact that all integers from $n$ onwards are red. Hence no such colouring exists.
A similar argument shows that there is no colouring meeting the given conditions other than (d), (e) and (f) in which 1 is coloured green.