Question Number 169727 by Mastermind last updated on 07/May/22
Commented by Mastermind last updated on 07/May/22
$${Okay},\:{thanks} \\ $$
Commented by mr W last updated on 07/May/22
$${you}\:{can}\:{express}\:{any}\:{natural}\:{number} \\ $$$${in}\:{this}\:{way} \\ $$$${n}=\sqrt{{n}\left({n}−\mathrm{1}\right)+\sqrt{{n}\left({n}−\mathrm{1}\right)+\sqrt{{n}\left({n}−\mathrm{1}\right)+...}}} \\ $$
Commented by cortano1 last updated on 07/May/22
$$\sqrt{\mathrm{42}+\sqrt{\mathrm{42}+\sqrt{\mathrm{42}+\sqrt{\mathrm{42}+\sqrt{...}}}}}\:=\:{x} \\ $$$$\:\mathrm{42}+{x}\:=\:{x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{42}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{7}\right)\left({x}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{7}\:;\:{x}=−\mathrm{6}\:\left({rejected}\right) \\ $$
Commented by Mastermind last updated on 07/May/22
$${your}\:{solution}\:@{mr}\:{W} \\ $$
Commented by mr W last updated on 07/May/22
$${n}=\sqrt{{m}+\sqrt{{m}+\sqrt{{m}+...}}} \\ $$$${n}=\sqrt{{m}+{n}} \\ $$$${n}^{\mathrm{2}} ={m}+{n} \\ $$$$\Rightarrow{m}={n}^{\mathrm{2}} −{n}={n}\left({n}−\mathrm{1}\right) \\ $$$$\Rightarrow{n}=\sqrt{{n}\left({n}−\mathrm{1}\right)+\sqrt{{n}\left({n}−\mathrm{1}\right)+\sqrt{{n}\left({n}−\mathrm{1}\right)+...}}} \\ $$
Commented by Mastermind last updated on 07/May/22
$${Thanks} \\ $$