Question Number 211467 by mnjuly1970 last updated on 10/Sep/24
$$ \\ $$$$\:\:\:{Solve}\:{the}\:{following}\:{equation} \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:{x}\:\rfloor\:+\:\sqrt{{x}\:−\sqrt{{x}}\:}\:=\:\lfloor\:{x}+\:\frac{\mathrm{1}}{{x}}\:\rfloor \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:{x}\:\rfloor\:{is}\:{floor}\:{of}\:,\:\:{x}\:\:,\: \\ $$
Answered by Frix last updated on 10/Sep/24
$$\sqrt{{x}−\sqrt{{x}}}\in\mathbb{R}\:\Rightarrow\:{x}=\mathrm{0}\vee{x}\geqslant\mathrm{1} \\ $$$$\lfloor{x}\rfloor\in\mathbb{N}\wedge\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor\in\mathbb{N}\:\Rightarrow\:\sqrt{{x}−\sqrt{{x}}}\in\mathbb{N} \\ $$$$ \\ $$$$\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor−\lfloor{x}\rfloor=\sqrt{{x}−\sqrt{{x}}} \\ $$$$\mathrm{1}.\:\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor=\lfloor{x}\rfloor \\ $$$$\:\:\:\:\:\mathrm{impossible}\:\mathrm{because}\:\sqrt{{x}−\sqrt{{x}}}=\mathrm{0}\:\Rightarrow \\ $$$$\:\:\:\:\:{x}=\mathrm{0}\vee{x}=\mathrm{1}\:\Rightarrow\:\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor\neq\lfloor{x}\rfloor \\ $$$$\mathrm{2}.\:\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor=\lfloor{x}\rfloor+\mathrm{1} \\ $$$$\:\:\:\:\:\Rightarrow\:\sqrt{{x}−\sqrt{{x}}}=\mathrm{1}\:\Rightarrow\:{x}=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 10/Sep/24
$$\:\cancel{ } \\ $$