Question

determine all pairs (x,y) of intergers satisfying the equation

Answer 1

\[\huge 1+2^x+2^{2x+1}=y^2\]

Answer 2

so far i did this
\[2^x+2 * 2^{2x}=y^2-1=(y-1)(y+1)\]
@mukushla

Answer 3

let \[a=2^x,a+2a^2=(y-1)(y+1)\]

Answer 4

\[a(1+2a)=(y-1)(y+1)\]

Answer 5

\[2^x(1+2^{x+1})=(y+1)(y-1)\]
shows that one of the factors y±1 is
divisible by 2 but not by 4 and the other by 2^x−1 but not by 2^x; hence x >3.

Answer 6

u r on right track :)

Answer 7

since (y+1),(y-1) divisible by 2 but not by 4 and the other by \[2^{x−1}\] but not by \[2^x\]
we have
\[y=2^{x-1}m\pm 1\] where m is a constant

Answer 8

exactly :) this will be helpful

http://openstudy.com/users/mukushla#/updates/5041ab9ee4b0d97e4dc5b0c8

Answer 9

exactly woah thats so helpful mukush???? thanx a lot

Answer 10

is this diophantine or basic number theory

Answer 11

the method you did there is exactly wat i wanted