Find number of x such that $\{x\}+\{x^2\}=1,x\in(0,20). $Where $\{\}$ denotes fractional part.

Question

Find number of x such that $\{x\}+\{x^2\}=1,x\in(0,20).$Where $\{\}$ denotes fractional part.

I did it by lot of case work searching for some simpler solution

Answer 1

On the interval (0,1), the function increases only once, and by monotonicity and continuity has exactly one solution on that interval. For every interval afterwards, $\{x^2\}$ "resets" multiple times. Each time it does, the function drops from a number bigger than one to a number less than one, so the equation picks up a solution for every reset. On the interval $(k-1,k)$, there are $k^2-(k-1)^2 =2k-1$ continuous pieces, hence there are $2k-1$ solutions. So the number of solutions on the interval $(0,n)$ is

$$\sum_{k=1}^n (2k - 1) = n(n+1) - n = n^2$$

In this case the number of solutions is $400$.