Show that

For any real number x, let {x} = x −[x] denote the fractional part where [x] is the
usual floor function.
(a) Show that there exist infinitely many positive rationals x such that {x}+{x^2} = 0.99.
(b) Show that there are no positive rational x such that {x} + {x^2} = 1.

(a)

For any non-negative integer $\displaystyle k,$ observe that $\displaystyle (10k+1.3)^2=100k^2+26k+1.69.$

Thus, if $\displaystyle x=10k+1.3\quad(k\in\mathbb{Z},\ k\geqslant0),$ we have $\displaystyle \{x\}+\{x^2\}=0.3+0.69=0.99.$

There are infinitely many non-negative integers $\displaystyle k$ and hence infinitely many such rationals $\displaystyle x.$

(b)

Note that for any $\displaystyle x\in\mathbb{R}^+,\ \{x\}=0\ \Leftrightarrow\ x\in\mathbb{Z}.$ Hence, in order to have $\displaystyle \{x\}+\{x^2\}=1,$ we must have $\displaystyle x\notin\mathbb{Z}.$ If $\displaystyle x$ is a positive non-integer rational, $\displaystyle x^2$ would have more decimal places than $\displaystyle x.$ This would mean that their fractional parts cannot add up to 1. Hence there are no positive rationals $\displaystyle x$ such that $\displaystyle \{x\}+\{x^2\}=1,$

Remark: There do however exist positive irrational numbers $\displaystyle x$ with $\displaystyle \{x\}+\{x^2\}=1,$ To see this, note that $\displaystyle f(x)=\{x\}+\{x^2\}$ is continuous on the interval $\displaystyle [0.1\,,\,0.9]$ and $\displaystyle f(0.1)<1$ while $\displaystyle f(0.9)>1;$ the result follows by the intermediate-value theorem.