Question Number 100951 by bemath last updated on 29/Jun/20

Commented byRasheed.Sindhi last updated on 29/Jun/20

By john santu(answer of  q#97746) Same as above.  2^x (1+2^(x+1) ) = (y−1)(y+1)  show that the factors y−1 and y+1  are even, exactly one of them   divisible by 4. Hence x≥3 and one  of these factors is divisible by 2^(x−1)   but not by 2^x . so y = 2^(x−1) m+ε , m odd , ε = ±1  plugging this into the original   equation we obtain 2^x (1+2^(x+1) )=(2^(x−1) m+ε)^2 −1  = 2^(2x−2) m^2 +2^x mε  or equivalently 1+2^(x+1) =2^(x−2) m^2 +mε  ⇔ thus we have the complete  list solutions (x,y) : (0,2),(0,−2),  (4,23),(4,−23).

Commented byRasheed.Sindhi last updated on 30/Jun/20

Sir bemath,  Your solution (which you have  given as a question) is solution  of your q#97746 which was  answered there by  sirjohn santu  and your this solution is same  as his. Why you′ve repeated it  and gave answer without question!!!  (Only to satisfy my curosity) :)

Commented by1549442205 last updated on 01/Jul/20

I guess perhaps he want everyone know above  his question with a  good solution for it

Commented byRasheed.Sindhi last updated on 01/Jul/20

May be.