2026/053) If both x and y are integers, how many pairs of solutions are there of the equation $(x−8)(x−10)=2^y$

As RHS is power of 2 so both $x-8$ and $ x-10$ should be power of 2 or they should be -ve of power or 2 because there are two terms on the left hand side so product shall be positive.

As $(x-8) - (x-10) = 2$ so the 2 values should be 4 and 2 or -2 and 4( because 2 and 4 the power of 2 which differ by 2)

Naturally x-8 shall be lower value

So taking $x -8 = 4$  or $x = 12$ we get $2^y= 4* 2 = 8$ or $y = 3$ giving $((x,y) = (12,3)$

Taking $x-8 = -2$ we get $x = 6$ and $y = 3$ or $((x,y) = (6,3)$

 So we have 2 set of solutions $(x,y) \in \{(12,3),(6,3)\}$