Question Number 58409 by Tawa1 last updated on 22/Apr/19
$$\mathrm{Prove}\:\mathrm{without}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\mathrm{expression}\:\:\:\left(\mathrm{1}\:+\:\sqrt{\mathrm{2}}\right)^{\mathrm{2n}} \:+\:\left(\mathrm{1}\:−\:\sqrt{\mathrm{2}}\right)^{\mathrm{2n}} \:\:\mathrm{is}\:\mathrm{even}\:\mathrm{for}\:\mathrm{every} \\ $$$$\mathrm{natural}\:\mathrm{number}\:\:\mathrm{n}. \\ $$
Commented by maxmathsup by imad last updated on 22/Apr/19
$${let}\:{A}_{{n}} =\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}} \:+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}} \:\:\Rightarrow{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\left(\sqrt{\mathrm{2}}\right)^{{k}} \:+\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\left(−\sqrt{\mathrm{2}}\right)^{{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\left\{\:\left(\sqrt{\mathrm{2}}\right)^{{k}} \:+\left(−\sqrt{\mathrm{2}}\right)^{{k}} \right\}\:=\sum_{{k}\:=\mathrm{2}{p}} \:\:\left(…\right)\:+\sum_{{k}=\mathrm{2}{p}+\mathrm{1}} \left(…\right) \\ $$$$=\sum_{{p}=\mathrm{0}} ^{{n}} \:\:{C}_{\mathrm{2}{n}} ^{\mathrm{2}{p}} \:\:\:\mathrm{2}^{{p}+\mathrm{1}} \:+\mathrm{0}\:=\mathrm{2}\:\left\{\sum_{{p}=\mathrm{0}} ^{{n}} \:\:{C}_{\mathrm{2}{n}} ^{\mathrm{2}{p}} \:\:\mathrm{2}^{{p}} \:\right\}\:\:\Rightarrow{A}_{{n}} {is}\:{even}\:. \\ $$
Commented by Tawa1 last updated on 23/Apr/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\:\mathrm{i}\:\mathrm{will}\:\mathrm{check}\:\mathrm{if}\:\mathrm{i}\:\mathrm{can}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{solution} \\ $$
Commented by maxmathsup by imad last updated on 23/Apr/19
$${you}\:{are}\:{welcome}\:{sir},\:{here}\:{i}\:{have}\:{used}\:{the}\:{binome}\:{formulae}\:. \\ $$