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Question Number 8025 by Nayon last updated on 28/Sep/16
find factor of ≫ (2^(4n+2) +1) at the same way expand 2^(58) +1
$${find}\:{factor}\:{of}\:\gg \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{2}^{\mathrm{4}{n}+\mathrm{2}} +\mathrm{1}\right) \\ $$$${at}\:{the}\:{same}\:{way}\:{expand}\:\mathrm{2}^{\mathrm{58}} +\mathrm{1} \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 28/Sep/16
2^(4n+2) +1 (2^(2n+1) )^2 +(1)^2 (2^(2n+1) )^2 +2(2^(2n+1) )(1)+(1)^2 −2(2^(2n+1) )(1) (2^(2n+1) +1)^2 −2^(2n+2) (2^(2n+1) +1)^2 −(2^(n+1) )^2 (2^(2n+1) +1−2^(n+1) )(2^(2n+1) +1+2^(n+1) ) (2^(2n+1) −2^(n+1) +1)(2^(2n+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)
$$\:\:\mathrm{2}^{\mathrm{4}{n}+\mathrm{2}} +\mathrm{1} \\ $$$$\:\:\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \right)\left(\mathrm{1}\right)+\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \right)\left(\mathrm{1}\right) \\ $$$$\:\:\:\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \\ $$$$\:\:\:\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{2}^{{n}+\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}−\mathrm{2}^{{n}+\mathrm{1}} \right)\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}+\mathrm{2}^{{n}+\mathrm{1}} \right) \\ $$$$\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{2}^{\mathrm{58}} +\mathrm{1}=\left(\mathrm{2}^{\mathrm{29}} \right)^{\mathrm{2}} +\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}^{\mathrm{29}} \right)^{\mathrm{2}} +\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}^{\mathrm{29}} \right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}^{\mathrm{29}} \right)\left(\mathrm{1}\right)+\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}^{\mathrm{29}} \right)\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}^{\mathrm{29}} +\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{2}^{\mathrm{15}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}^{\mathrm{29}} +\mathrm{1}−\mathrm{2}^{\mathrm{15}} \right)\left(\mathrm{2}^{\mathrm{29}} +\mathrm{1}+\mathrm{2}^{\mathrm{15}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}^{\mathrm{29}} −\mathrm{2}^{\mathrm{15}} +\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{29}} +\mathrm{2}^{\mathrm{15}} +\mathrm{1}\right) \\ $$
Answered by prakash jain last updated on 02/Oct/16
answer in comments
$$\mathrm{answer}\:\mathrm{in}\:\mathrm{comments} \\ $$

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