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Put y = [x], z = {x} now y is an integer, z is in [0, 1) we have yz = 2005y + 2005z y = 2005z / (z - 2005) = 2005 - 2005^2/(z - 2005) So z - 2005 divides 2005^2 ; in particular, for some integer k, z = 2005(2005 + k)/k. If k > 0 then $\frac{2005+k}{k} > 1$ and z > 1. If -2004<k<0 then $z < 0$. If k = -2005, -2006 we get that z is 0, 2005/2006. If k < -2006 then it is increasing towards 2005, starting at 2(2005)/2007 > 1. From z = 0 we get x = 0 easily. From z = 2005/2006, we get y = -1 easily.

$[x]\{x\}=2005[x]+2005\{x\}$ so $\{x\}=\frac {2005[x]}{[x]-2005} $and the only condition is $0\leq \{x\}<1$ now if $[x]<2005$ then $[x]-2005<0$ so $2005[x]\leq 0$ so $[x]\leq 0$ now let $n=-[x]$ so we have $0<\frac{2005n}{n+2005}<1$ so $2005n<2005+n$ so $n<\frac{2005}{2004}$ so $n=0$ or $n=1$ so $[x]=-1$ or $[x]=0$. for $[x]>2005$ you obtain $[x]<\frac{-2005}{2004}$ so no solution. the only solutions are then $x=\frac{-1}{2006}$ and $x=0$

singular posted his solution while i was writing mine so that's why there are two solutions

we can get equation [x]=-2005*{x}/(2005-{x}) so we have 2005>=2005 - {x} >2004 and 2005>2005*{x} so [x] can be 1,0,-1 [x]=0 -> {x}=0 -> x=0 [x]=-1 -> {x}=2005/2006 -> x=-1/2006 and [x]=1 -> {x}=-2005/2004 contradiction... solutions x=0, x=-1/2006

let [x]=a,{x}=b ab=2005(a+b) add 2005^2 both sides we get (a-2005)(b-2005)=2005^2 since a-2005 not 0.we get b-2005=2005^2/(a-2005) <=> b=2005a/(a-2005)<1 if a>0 we get 2004a+2005<0 contr. if a<0 let a=-c =>2004c<2005 =>c=1=>a=-1 b=2005/2006 =>x=-1/2006 if a=0 then b=0=>x=0 x=0,-1/2006