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Question Number 174960 by alcohol last updated on 14/Aug/22
$${solve}\:{for}\:{all}\:{x}\: \\ $$$$\lfloor{x}^{\mathrm{2}} \rfloor\:−\:\lfloor{x}\rfloor^{\mathrm{2}} \:=\:\mathrm{100} \\ $$
Answered by mr W last updated on 15/Aug/22
$${say}\:{x}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\lfloor{n}^{\mathrm{2}} +\mathrm{2}{nf}+{f}^{\mathrm{2}} \rfloor−{n}^{\mathrm{2}} =\mathrm{100} \\ $$$$\lfloor\mathrm{2}{nf}+{f}^{\mathrm{2}} \rfloor=\mathrm{100} \\ $$$$\mathrm{100}=\lfloor\mathrm{2}{nf}+{f}^{\mathrm{2}} \rfloor<\mathrm{2}{n}+\mathrm{1} \\ $$$$\Rightarrow{n}>\mathrm{49}.\mathrm{5}\:\Rightarrow{n}\geqslant\mathrm{50} \\ $$$$\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{100}+{n}^{\mathrm{2}} \\ $$$$\mathrm{100}+{n}^{\mathrm{2}} \leqslant{x}^{\mathrm{2}} <\mathrm{101}+{n}^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{100}+{n}^{\mathrm{2}} }\leqslant{x}<\sqrt{\mathrm{101}+{n}^{\mathrm{2}} }\:{with}\:{n}\geqslant\mathrm{50} \\ $$
Commented by alcohol last updated on 15/Aug/22
$${thank}\:{you} \\ $$$${please}\:{can}\:{you}\:{explain}? \\ $$
Commented by mr W last updated on 15/Aug/22
$${i}\:{thought}\:{i}\:{have}\:{given}\:{all}\:{necessary} \\ $$$${steps}. \\ $$$${where}\:{do}\:{you}\:{need}\:{more}\:{explanation}? \\ $$
Commented by alcohol last updated on 15/Aug/22
$${i}\:{have}\:{no}\:{idea}\:{about}\:{it}.\:{if}\:{you}\:{can}\:{propose}\:{me}\:{a}\:{book}\: \\ $$$${to}\:{read}\:{i}\:{will}\:{be}\:{grateful} \\ $$
Commented by mr W last updated on 15/Aug/22
$${since}\:{you}\:{asked}\:{this}\:{question},\:{i}\:{assumed} \\ $$$${that}\:{you}\:{know}\:{what}\:{the}\:{greatest}\:{integer} \\ $$$${function}\:\lfloor{x}\rfloor\:{is}.\:{i}\:{don}'{t}\:{know}\:{any}\:{books} \\ $$$${about}\:{such}\:{things}. \\ $$