We have
2x + 1 = ny ... 1
2y + 1 = mx .... 2 with m and n > 0. Without loass of generality let m >= n
from the above (1) and(2) we have n,y,m and x all are odd
2x + 1 = ny ... 1
2y + 1 = mx .... 2 with m and n > 0. Without loass of generality let m >= n
from the above (1) and(2) we have n,y,m and x all are odd
Again from (1)
4x + 2 = 2ny = n(mx-1) as 2y = mx-1 from (2)
or 4x + 2 = nmx - n
or x(nm-4 ) = n+ 2
or x= (n+2)/(nm- 4)
4x + 2 = 2ny = n(mx-1) as 2y = mx-1 from (2)
or 4x + 2 = nmx - n
or x(nm-4 ) = n+ 2
or x= (n+2)/(nm- 4)
now as x n + 2 >= nm - 4 or n(m-1) <= 6 and nm > 4
n =1 => m= 5 or 7
n =1 => m= 5 or 7
n =3 => m = 3
n= 1, m= 5 => x = 1 and y = 3
n =1,m =7 => x = 3, y = 7
n= 3 , m= 3 => x = 1 , y =1
So solution set (1,1) , (1,3) (3,7)
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