find-factor-of-2-4n-2-1-at-the-same-way-expand-2-58-1-

Question Number 8025 by Nayon last updated on 28/Sep/16

f i n d f a c t o r o f ≫

(2^{4 n + 2} + 1)

a t t h e s a m e w a y e x p a n d 2^{58} + 1

Commented by Rasheed Soomro last updated on 28/Sep/16

2^{4 n + 2} + 1

{(2^{2 n + 1})}^{2} + {(1)}^{2}

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

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

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

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

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

2^{58} + 1 = {(2^{29})}^{2} + {(1)}^{2}

= {(2^{29})}^{2} + {(1)}^{2}

= {(2^{29})}^{2} + 2 (2^{29}) (1) + {(1)}^{2} - 2 (2^{29}) (1)

= {(2^{29} + 1)}^{2} - {(2^{15})}^{2}

= (2^{29} + 1 - 2^{15}) (2^{29} + 1 + 2^{15})

= (2^{29} - 2^{15} + 1) (2^{29} + 2^{15} + 1)

Answered by prakash jain last updated on 02/Oct/16

answer in comments