WakeUp wrote: Every integer is to be coloured blue, green, red, or yellow. Can this be done in such a way that if $a, b, c, d$ are not all $0$ and have the same colour, then $3a-2b = 2c-3d$? Obviously not. Choose $a=b=c=d=1$ and you get a contradiction. And obviously, the word "different" is not missing, since the statement says "not all zero" which means that it's allowed to have two of them zero and the two others non zero, and so having some of $a,b,c,d$ identical is allowed.

pco wrote: Obviously not. Choose $a=b=c=d=1$ and you get a contradiction. Those values of $a, b, c$ and $d$ do not satisfy $3a-2b=2c-3d$

SKhan wrote: pco wrote: Obviously not. Choose $a=b=c=d=1$ and you get a contradiction. Those values of $a, b, c$ and $d$ do not satisfy $3a-2b=2c-3d$ Sure, that's why I choosed them : obviously they have the same colour but dont match the equality $3a-2b=2c-3d$. Hence the conclusion : It's impossible to do this colorization.

pco wrote: SKhan wrote: pco wrote: Obviously not. Choose $a=b=c=d=1$ and you get a contradiction. Those values of $a, b, c$ and $d$ do not satisfy $3a-2b=2c-3d$ Sure, that's why I choosed them : obviously they have the same colour but dont match the equality $3a-2b=2c-3d$. Hence the conclusion : It's impossible to do this colorization. Yeah, I just misread the question a bit. Sorry.