Question-195412

Question Number 195412 by Calculusboy last updated on 02/Aug/23

Answered by AST last updated on 02/Aug/23

v_2 (a^2^n −1)=v_2 (a−1)+n+v_2 (a+1)−1≥n+2 ⇒2^(n+2) ∣a^2^n −1

$${v}_{\mathrm{2}} \left({a}^{\mathrm{2}^{{n}} } −\mathrm{1}\right)={v}_{\mathrm{2}} \left({a}−\mathrm{1}\right)+{n}+{v}_{\mathrm{2}} \left({a}+\mathrm{1}\right)−\mathrm{1}\geqslant{n}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}^{{n}+\mathrm{2}} \mid{a}^{\mathrm{2}^{{n}} } −\mathrm{1} \\ $$

Commented by Calculusboy last updated on 02/Aug/23

$${thanks} \\ $$

Answered by MM42 last updated on 02/Aug/23

proof by induction 1)n=1⇒8∣a^2 −1=(2m+1)^2 −1=4m(m+1)=8k ✓ 2)n=k → 2^(n+2) ∣a^2^n −1⇔a^2^n −1=2^(n+2) ×t 3)2^(n+3) ∣a^2^(n+1) −1 ⇔ a^2^(n+1) −1=2^(n+3) ×u ? a^2^(n+1) −1=(a^2^n −1)(a^2^n +1) =2^(n+2) ×t×2t′=2^(n+3) ×u ✓

$${proof}\:{by}\:{induction} \\ $$$$\left.\mathrm{1}\right){n}=\mathrm{1}\Rightarrow\mathrm{8}\mid{a}^{\mathrm{2}} −\mathrm{1}=\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{4}{m}\left({m}+\mathrm{1}\right)=\mathrm{8}{k}\:\checkmark \\ $$$$\left.\mathrm{2}\right){n}={k}\:\rightarrow\:\mathrm{2}^{{n}+\mathrm{2}} \:\mid{a}^{\mathrm{2}^{{n}} } −\mathrm{1}\Leftrightarrow{a}^{\mathrm{2}^{{n}} } −\mathrm{1}=\mathrm{2}^{{n}+\mathrm{2}} ×{t} \\ $$$$\left.\mathrm{3}\right)\mathrm{2}^{{n}+\mathrm{3}} \:\mid{a}^{\mathrm{2}^{{n}+\mathrm{1}} } \:−\mathrm{1}\:\Leftrightarrow\:{a}^{\mathrm{2}^{{n}+\mathrm{1}} } −\mathrm{1}=\mathrm{2}^{{n}+\mathrm{3}} ×{u}\:? \\ $$$${a}^{\mathrm{2}^{{n}+\mathrm{1}} } \:−\mathrm{1}=\left({a}^{\mathrm{2}^{{n}} } −\mathrm{1}\right)\left({a}^{\mathrm{2}^{{n}} } +\mathrm{1}\right)\:\:\: \\ $$$$=\mathrm{2}^{{n}+\mathrm{2}} ×{t}×\mathrm{2}{t}'=\mathrm{2}^{{n}+\mathrm{3}} ×{u}\:\checkmark \\ $$

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