2021/064) Solve in positive integer a,b $a^3-b^3 = ab + 11$

This type of problem is best solved by finding the bound and checking values in the bound

We have a=b is not a solution because LHS = 0 and RHS= 11

So $(a-b) >=1$ as $a > b$ because RHS is positive

Now $a^3-b^3 = (a-b)(a^2 + b^2 + ab)$

As and b are positive so AM, GM inequality gives   $a^2 + b^2 >= 2ab$

So we have $a^3 - b^3 >=  (a-b)(3ab)$

Or $a^3-b^3 >= 3ab$

So from the given relation $3ab < = ab +11 $ or $ 2ab <=11$ or $ab < 6 $ so $b<=2$ as   $a<=b$

b = 1 gives $a^3 = a + 12$ a =3 gives LHS = 8 RHS = 15

a = 4 gives LHS = 64 RHS = 16 so no integer solution LHS is greater than RHS 

b = 2 gives $a^3 = 2a + 11$ a =3 given LHS = 8, RHS = 17

a = 4 gives LHS = 64 RHS = 19 

So no solution

Hence there is no solution to this equation