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.