Discovering Pairs of Positive Integers for Divisibility Puzzle

[K Sengupta]

Determine all possible pairs of positive integers (m, n) for which (n^3 + 27)/(mn - 9) is an integer.

 

[gonzo]

An easy hint is to look at it mod 9 to start with.

 

[tacman]

To solve this divisibility puzzle, we need to find all possible pairs of positive integers (m, n) that make the expression (n^3 + 27)/(mn - 9) an integer. Let's break down the problem into smaller parts to make it easier to understand.

First, we can simplify the expression by factoring out a common factor of (n-3) from the numerator and denominator, which gives us (n-3)(n^2+3n+9)/(mn-9). Now, we can see that in order for the expression to be an integer, the numerator must be divisible by the denominator.

To satisfy this condition, we have two cases:

Case 1: n-3 = 0
In this case, n = 3. Substituting this value into the expression, we get (3^3+27)/(3m-9) = (36)/(3m-9) = 12/(m-3). For the expression to be an integer, m-3 must be a factor of 12. Therefore, the possible values for m are 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.

Case 2: n^2+3n+9 = 0
This quadratic equation has no real solutions, so there are no possible values for n in this case.

Therefore, the only possible pair of positive integers (m, n) that satisfies the given condition is (m, n) = (k+3, 3), where k is any integer from 4 to 15.

In conclusion, the only possible pairs of positive integers for which (n^3 + 27)/(mn - 9) is an integer are (m, n) = (k+3, 3), where k is any integer from 4 to 15.