Question Number 167367 by mnjuly1970 last updated on 14/Mar/22
Commented by mr W last updated on 15/Mar/22
$${for}\:{a}=\mathrm{2}\:{there}\:{is}\:{no}\:{solution}.\:{for}\:{any} \\ $$$${other}\:{values}\:{of}\:{a}>\mathrm{1}\:{there}\:{is}\:{always} \\ $$$${solution},\:{i}\:{think}. \\ $$
Answered by TheSupreme last updated on 15/Mar/22
$$\mathrm{0}<{x}<\mathrm{1}\:\lfloor\left({a}^{\mathrm{2}} −{a}\right){x}\rfloor=\mathrm{2} \\ $$$$\mathrm{1}\leqslant{x}<\mathrm{2}\:\lfloor\left({a}^{\mathrm{2}} −{a}\right){x}\rfloor=\mathrm{1} \\ $$$${x}\geqslant\mathrm{2}\:{no}\:{solution} \\ $$