Question Number 73308 by wo1lxjwjdb last updated on 10/Nov/19
$${what}\:{are}\:{the}\:{solutions} \\ $$$${of}\:\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}}={n}\:{where}\:{n}\in\mathbb{N} \\ $$
Commented by MJS last updated on 10/Nov/19
$${n}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{k}} +\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{k}} \right)=\mathrm{cosh}\:\left({k}\mathrm{ln}\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right) \\ $$$${x}_{{k}} =−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sinh}\:\left({k}\mathrm{ln}\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right) \\ $$$${k},\:{n}_{{k}} ,\:{x}_{{k}} \:\in\mathbb{N} \\ $$
Commented by MJS last updated on 10/Nov/19
$$\mathrm{you}\:\mathrm{mean}\:{n}\in\mathbb{N} \\ $$
Commented by kaivan.ahmadi last updated on 10/Nov/19
$${n}\in\mathbb{N} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}={n}^{\mathrm{2}} \Rightarrow\mathrm{3}{x}^{\mathrm{2}} ={n}^{\mathrm{2}} −\mathrm{1}\Rightarrow{x}^{\mathrm{2}} =\frac{{n}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}}\Rightarrow \\ $$$${x}=\sqrt{\frac{{n}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}}} \\ $$
Commented by wo1lxjwjdb last updated on 10/Nov/19
$${when}\:{is}\:{it}\:{a}\:{whole}\:{number}? \\ $$
Commented by MJS last updated on 10/Nov/19
$${n}\in\mathbb{N} \\ $$$${x}\in? \\ $$$$\mathrm{if}\:{x}\in\mathbb{R} \\ $$$${x}=\pm\sqrt{\frac{{n}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}}};\:{n}>\mathrm{0} \\ $$$$\:\:\:\:\:\left[\mathrm{for}\:{x}\in\mathbb{C}\:\mathrm{we}\:\mathrm{also}\:\mathrm{get}\:{n}=\mathrm{0};\:{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{i}\right] \\ $$$$\mathrm{if}\:{x}\in\mathbb{N} \\ $$$$\mathrm{same}\:\mathrm{equation},\:\mathrm{solutions}\:\mathrm{are}: \\ $$$${x}\in\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{4},\:\mathrm{15},\:\mathrm{56},\:\mathrm{209},\:\mathrm{780},\:…\right\} \\ $$$${n}\in\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{7},\:\mathrm{26},\:\mathrm{97},\:\mathrm{362},\:\mathrm{1351},\:…\right\} \\ $$
Commented by kaivan.ahmadi last updated on 10/Nov/19
$${Mr}\:{MJS}\:{answered}\:{this}\:{quoestion}.\: \\ $$