**Problem 7 of the Canadian Mathematical Olympiad 1969**

Show that there are no integers a, b, c for which a² + b² - 8c = 6.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to my wife Quỳnh Châu)**

Adding 2ab to both sides, we have (a + b)² = 2(ab + 4c + 3) so that a + b is even, say a + b = 2d. It then follows that ab + 4c + 3 = 2d² where d is an integer. Since 2d² is even, the product ab must be odd, implying that both a and b have to be odd.

Now let a = 2m +1 and b = 2n + 1 where m and n are integers. Substituting these into the original equation, we successively have

(2m + 1)² + (2n + 1)² = 2 (4c + 3)

4m² + 4m + 4n² + 4n + 2 = 2 (4c + 3)

m² + m + n² + n = 2c + 1

m(m + 1) + n(n + 1) = 2c + 1 (*)

Now note that the product of two consecutive integers is always even since one of them is necessarily even. Therefore, the sum on the left of (*) is even whereas the one on the right is odd. The assumption that both a and b are odd leads to a contradiction. Therefore, one cannot find integers for a, b and c to satisfy the equation, as claimed.

**Solution #2**

Assume a² + b² - 8c = 6, then a² + b² = 8c + 6, meaning that a² + b² is even. This is only possible if a and b are of the same parity. If both are even, say a = 2n and b = 2m, then 4n² + 4m² - 8c = 6. This could not be as the left hand side is divisible by 4 whereas the right hand side is only divisibly by 6. Assume both a and b are odd: a = 2n + 1, b = 2m + 1. Then 4n(n + 1) + 4m(m + 1) = 8c + 4, or n(n + 1) + m(m + 1) = 2c + 1. However, the left hand side is the sum of two even numbers, whereas the right hand side is odd. Another contradiction.