RagvaloD wrote: Find all integer $b$ such that $[x^2]-2012x+b=0$ has odd number of roots. Let $\lfloor x^2\rfloor =n$ and we get $x=\frac{n+b}{2012}$ And so we are looking for $n$ such that $n+1>\frac{(n+b)^2}{2012^2}\ge n$ Which is $2012^2-b^2>n^2+(2b-2012^2)n\ge -b^2$ From there, we see that $n$ solution implies $2012^2-2b-n$ solution too. So, in order to have an odd number of solution, we need $\frac{2012^2}2-b$ solution. And so $2012^2-b^2>\left(\frac{2012^2}2-b\right)^2+(2b-2012^2)\left(\frac{2012^2}2-b\right)\ge -b^2$ Which is $\frac{2012^2}4+1>b\ge\frac{2012^2}4$ And so a unique possibility $\boxed{b=\frac{2012^2}4}$