Find all pairs of integers (x, y) satisfying 3xy – x – 2y = 8.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to thi sĩ Kim Ngân)**

Let x = y + z where z is an integer. The equation can be written as 3y² +3(z – 1)y – z – 8 = 0.

Solving for y, we have y_{1,2}= \frac {1}{6} (3 – 3z ± \sqrt {9z² - 6z + 105} ) (*)

Now assume 9z² - 6z + 105 is a square, a requirement for y to be an integer. Let

9z² - 6z + 105 = n², where n is an integer, or 9z² - 6z + 105 - n² = 0.

Solving for z, we have z_{1,2} = \frac {1}{9} [3 ± \sqrt {9(n² - 104)} ]

So now n² - 104 has to be a square. Let n² - 104 = m² (**) where m is also an integer.

To satisfy (**), both n and m have to be either odd or even.

a) For the case of both n and m being even integers, let n = 2p and m = 2q,

where p and q are both integers, and now (**) becomes 4p² – 104 = 4q², or (p - q) (p + q) = 26, and p – q and p + q can take on these values (p – q, p + q) = (1, 26), (2, 13), (13, 2) or (26, 1)

or 2p = 27 or 2p = 15 and neither of these is possible because 2p is an even integer.

b) If they are both odd integers, let n = 2p + 1 and m = 2q + 1,

where p and q are both integers. Now (**) becomes 4p² + 4p – 104 = 4q² + 4q, or (p - q) (p + q + 1) = 26, and p – q and p + q + 1 can take on these values

(p – q, p + q + 1) = (1, 26), (2, 13), (13, 2) or (26, 1) or p = 13, p = 7 which make n = 27 or n = 15, respectively.

For n = 27, z_{1,2} = 78/9 or -8 for which z = –8 is the only integer solution.

Substituting z = -8 into (*), we have y = 0 and y = 9, and subsequent substitution of these values of y into the equation of the problem, we have y = 0, x = -8 and y = 9, x = 1

For n = 15, z_{1,2} = 4 or -30/9 for which z = 4 is the only integer solution.

Substituting z = 4 into (*), we have y = -4 and y = 1, and subsequent substitution of these values of y into the equation of the problem, we have y = -4, x = 0 and y = 1, x = 5

Answers: All pairs of integers (x, y) satisfying 3xy – x – 2y = 8 are

(x, y) = (-8, 0), (0, -4), (1, 9) and (5, 1)