Question Number 155978 by mnjuly1970 last updated on 06/Oct/21
Commented by mr W last updated on 06/Oct/21
$${i}\:{think}\:{for}\:{any}\:{a}\:{the}\:{eqn}.\:{has}\:{always} \\ $$$${solution}. \\ $$
Commented by mr W last updated on 07/Oct/21
$${i}\:{thought}\:{about}\:{the}\:{question}\:{again}. \\ $$$${i}\:{think}\:\lfloor\mathrm{2}{x}\rfloor+\lfloor{x}\rfloor=\mathrm{2}\:{has}\:{no}\:{solution}, \\ $$$${therefore} \\ $$$${a}^{\mathrm{2}} −{a}\neq\mathrm{2} \\ $$$${a}^{\mathrm{2}} −{a}−\mathrm{2}\neq\mathrm{0} \\ $$$$\left({a}−\mathrm{2}\right)\left({a}+\mathrm{1}\right)\neq\mathrm{0} \\ $$$${a}\neq\mathrm{2},\:{a}\neq−\mathrm{1} \\ $$$${is}\:{this}\:{the}\:{correct}\:{answer}\:{sir}? \\ $$