Let a and b natural numbers such that 2 a-b, a-2 b and a+b are all distinct squares. What is the smallest possible value of b ?

Step-by-step solution

2 a - b = k ₁^2 ......(1) a-2 b=k₂^2 ......(2) a+b=k₃ ^2 ......(3) Add (2) & (3) we get 2 a - b = k ₂^2+ k ₃^2 k ₂^2+ k ₃^2= k ₁^2 ( k ₂ < k ₃ ) For least ' b ' difference of k ₃ ^2 & k ₂ ^2 is also least and must be multiple of 3 k ₂^2= a -2 ~b = a ^2 & k ₃^2= a + b =12^2 k ₃^2- k ₂^2=3 ~b =144-81=63 ~b =21 least b is 21