Prove that 41 divides 2^(20) −1


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Question Number 35004 by Rasheed.Sindhi last updated on 14/May/18

$Prove that 41 divides 2^{20} - 1$
	Answered by tanmay.chaudhury50@gmail.com last updated on 14/May/18

	$l e t u s s o l v e a n a l y t i c a l l y . . .$ $2^{20} - 1 = {(2^{5})}^{4} - 1$ ${(32)}^{4} - 1 = {(41 - 9)}^{4} - 1$ $= {(_{} 41)}^{4} - 4 C_{1} {(41)}^{3} 9 + 4 C_{2} {(41)}^{2} . 9^{2} - 4 C_{3} (41) . 9^{3}$ $+ 9^{4} - 1$ $s o e x c e p t (9^{4}) r e s t t e r m s a r e d i v i s i b l e b y 41$ $n o w 9^{4} - 1$ ${(9^{2})}^{2} - 1$ ${(82 - 1)}^{2} - 1$ ${(82^{})}^{2} - 2 \times 82 \times 1 + 1 - 1$ ${(82)}^{2} - 2 \times 82 \times 1 t h i s t w o t e r m s a l s o d i v d i b l e$ $b y 41 s i n c e 82 i s d i v i s i b l e b y 41$
	Commented by Rasheed.Sindhi last updated on 15/May/18

	$G^{\overset{⌢}{O} \overset{⌢}{O}} D A p p r o a c h!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 15/May/18

	$t h a n x . . . w e c a n p o s t q u e s t i o n a l o n g w i t h$ $f i n a l a n s w e r s o t h a t w e g e t s a t i s f a c t i o n . . .$ $t h e p e o p l e w h o p o s t q u e s t i o n m a y p o s t q u e s t i o n$ $d i f f e r e n t v a r i e t y . . .$
	Answered by MJS last updated on 14/May/18

	$2^{20} \mod 41 = 1$ $(2^{20} \mod 41) = {(2^{10} \mod 41)}^{2} = {(2^{5} \mod 41)}^{4} =$ $= (- 9) (- 9) (- 9) (- 9) = 81 \times 81 = (- 1) (- 1) = 1$ $multiplying remainders can be great fun; -)$
	Commented by Rasheed.Sindhi last updated on 15/May/18

	$V_{r}^{e} y N i c e! . . .$ $But using congruence notation were$ $easier and prettier, I think .$
	Commented by MJS last updated on 14/May/18

	$true$
	Answered by ajfour last updated on 14/May/18

	$41 = 2^{5} + 3^{2}$ $2^{20} = {(41 - 3^{2})}^{4}$ $= 41^{4} - 4 . 3^{2} . 41^{3} + 6 . 3^{4} . 41^{2}$ $- 4 . 3^{6} . 41 + 3^{8}$ $3^{8} = {(2 \times 41 - 1)}^{2} = 4 \times 41^{2} - 4 \times 41 + 1$ $h e n c e 2^{20} = k (41) + 1 (\forall k \in N)$ $o r 41 d i v i d e s 2^{20} - 1 .$
	Commented by Rasheed.Sindhi last updated on 15/May/18

	$G^{\overset{⌢}{O} \overset{⌢}{O}} D A p p r o a c h!$
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