Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 163259 by SANOGO last updated on 05/Jan/22

	Answered by TheSupreme last updated on 05/Jan/22

	$1)$ $b^{n} - 1 \equiv 0 (b - 1)$ $(b - 1) P_{n - 1} (b) = b^{n} - 1$ $P_{n - 1} = \underset{i = 0}{\sum^{n - 1}} b^{i}$ $(b - 1) \underset{i = 0}{\sum^{n - 1}} b^{i} = \underset{i = 0}{\sum^{n - 1}} b^{i + 1} - b^{i} = b^{n} - 1$ $2)$ $a^{k} - 1 i s p r i m e$ $a^{k} - 1 \equiv 1 (2)$ $a^{k} \equiv 0 (2)$ $a \equiv 0 (2)$ $2 t = a$ $2^{k} t^{k} - 1 i s p r i m e$ $a^{k} - 1 i s a l w a y s d i v i d e d b y a - 1$ $s o w e h a v e a - 1 = a^{k} - 1 o r a - 1 = 1$ $\forall p < a^{k} - 1 : a^{k} - 1 \equiv m (p) w i t h m \neq 0$
	Commented by SANOGO last updated on 06/Jan/22

	$m e r c i b i e n$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum