Question Number 189464 by Rupesh123 last updated on 17/Mar/23
Answered by Rasheed.Sindhi last updated on 17/Mar/23
$$\mathrm{4}{a}_{{n}+\mathrm{1}} ^{\mathrm{2}} −\mathrm{4}{a}_{{n}} {a}_{{n}+\mathrm{1}} +{a}_{{n}} ^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${n}=\mathrm{1}: \\ $$$$\mathrm{4}{a}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{4}{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{a}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{4}{a}_{\mathrm{1}} \pm\sqrt{\mathrm{16}−\mathrm{4}\left(\mathrm{4}\right)\left({a}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}\right)}\:}{\mathrm{2}\left(\mathrm{4}\right)} \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}{a}_{\mathrm{1}} \pm\mathrm{4}\sqrt{\mathrm{2}−{a}_{\mathrm{1}} ^{\mathrm{2}} }\:}{\mathrm{2}\left(\mathrm{4}\right)} \\ $$$$\:\:\:\:\:=\frac{{a}_{\mathrm{1}} \pm\sqrt{\mathrm{2}−{a}_{\mathrm{1}} ^{\mathrm{2}} }\:}{\mathrm{2}}\in\mathbb{Z}\Rightarrow{a}_{\mathrm{1}} =\pm\mathrm{1} \\ $$
Commented by Rupesh123 last updated on 18/Mar/23
Excellent!