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

\Rightarrow 2^{n + 2} ∣ a^{2^{n}} - 1

Commented by Calculusboy last updated on 02/Aug/23

t h a n k s

Answered by MM42 last updated on 02/Aug/23

p r o o f b y i n d u c t i o n

1) n = 1 \Rightarrow 8 ∣ a^{2} - 1 = {(2 m + 1)}^{2} - 1 = 4 m (m + 1) = 8 k ✓

2) n = k \to 2^{n + 2} ∣ a^{2^{n}} - 1 \Leftrightarrow a^{2^{n}} - 1 = 2^{n + 2} \times t

3) 2^{n + 3} ∣ a^{2^{n + 1}} - 1 \Leftrightarrow a^{2^{n + 1}} - 1 = 2^{n + 3} \times u ?

a^{2^{n + 1}} - 1 = (a^{2^{n}} - 1) (a^{2^{n}} + 1)

= 2^{n + 2} \times t \times 2 t^{'} = 2^{n + 3} \times u ✓

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