Q13/057) How many ordered pairs of integers (a,b) satisfy (1/a) + (1/b)= (1/6)

1/a + 1/b = 1/6
=> 6 * (b + a) = ab
=> 6b + 6a = ab
=> 6b = ab - 6a
=> a = 6b / (b - 6) = a So a, b and b-6 cannot be zero ( b-6 = 0 => b = 6 and a = infinite
letting b - 6 = u
we get a = 6 * (u + 6) / (u)
=> a = 6u/u + 36/u = a
=> a =6 + 36/u and b = u + 6
for a to be integer u has to be  integer and factor or 36 so we get the values as

u = 36( b = 42, a = 7), u = 18( b = 24, a =8  ), u = 12 ( b= 18, a = 9), u = 9( b= 15, a= 10), u = 6 ( b = 12, a = 12)

u = 4( b = 10, a = 15),u =3( b = 9, a = 18), u = 2( b= 8,  a = 24),u = 1 ( b = 7, a = 42)

u = -1 ( b = 5, a = - 30) ,u = -2 ( b = 4, a = - 12),u = -3 ( b = 3, a = - 6),u = - 4 ( b = 2, a = - 3),u = - 6( b = 0, a 0) inadmissible,u = - 9( b= -3, a = 2),u =- 12(b = 0 6, a = 3),u = - 18( b= - 12, a = 4),u = 36 (b = - 30, a= 6)

the above are solution there are 17 of them