LI OM - I - Task 5

Determine all pairs $ (a,b) $ of natural numbers for which the numbers $ a^3 + 6ab + 1 $ and $ b^3 + 6ab + 1 $ are cubes of natural numbers.
Let $ a $ and $ b $ be numbers satisfying the conditions of the problem. Without loss of generality, we can assume that $ a \leq b $. Then






Since the number $ b^3+6ab + 1 $ is a cube of an integer, the following equality must hold






Transforming the above relationship equivalently, we obtain step by step






It remains to determine for which pairs $ (a, b) $ satisfying condition (2) the number






is a cube of an integer. By virtue of relation (2), the above number is equal to $ a^3 + 12a^2 - 6a + 1 $. From the inequality






it follows that if the number $ a^3 + 12a^2 - 6a + 1 $ is a cube of an integer, then it is equal to $ (a+ 1)^3 $, $ (a + 2)^3 $, or $ (a + 3)^3 $. Therefore, we need to consider three cases:
(a) $ a^3 + 12a^2 - 6a + 1 = (a + 1)^3 $
In this case, we obtain $ 9a^2 - 9a = 0 $, from which $ a = 0 $ or $ a = 1 $.
(b) $ a^3 + 12a^2 - 6a + 1 = (a + 2)^3 $
Then we have $ 6a^2 - 18a - 7 = 0 $. The obtained equation has no integer solutions, as the left side is not divisible by $ 3 $, while the right side is.
(c) $ a^3 + 12a^2 - 6a + 1 = (a + 3)^3 $
Then we get $ 3a^2 - 33a - 26 = 0 $. Similarly to case (b), the obtained equation has no integer solutions, as the left side is not divisible by $ 3 $, while the right side is equal to $ 0 $.
Ultimately, we have found only one pair $ (a, b) = (1, 1) $ satisfying the conditions of the problem.