Math Question and Answers


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 195412 by Calculusboy last updated on 02/Aug/23

	Answered by deleteduser1 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

	$p r o o f b y i n d u c t i o n$ $1) n = 1 \Rightarrow 8 ∣ a^{2} - 1 = {(2 m + 1)}^{2} - 1 = 4 m (m + 1) = 8 k ✓$ $2) n = k \to 2^{n + 2} ∣ a^{2^{n}} - 1 \Leftrightarrow a^{2^{n}} - 1 = 2^{n + 2} \times t$ $3) 2^{n + 3} ∣ a^{2^{n + 1}} - 1 \Leftrightarrow a^{2^{n + 1}} - 1 = 2^{n + 3} \times u ?$ $a^{2^{n + 1}} - 1 = (a^{2^{n}} - 1) (a^{2^{n}} + 1)$ $= 2^{n + 2} \times t \times 2 t^{'} = 2^{n + 3} \times u ✓$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum