**Problem 5 of the Canadian Mathematical Olympiad 1972**

Prove that the equation x³ +11³ = y³ has no solution in positive integers x and y.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

Rearrange the equation to (x – y)³ + 3x²y – 3xy² = -11³ or

(y – x) [ (y – x)² + 3xy ] = 11³

Since 11 is a prime integer, the only possible values for y – x are 1, 11, 11² or 11³

If y – x = 1 then 3xy = 11³ -1 = 1330 or xy = 1330/3 which is not an integer and therefore this is not a possible scenario.

If y – x = 11 then 3xy = 0 and either x or y must be 0 and not positive as required.

If y – x = 11² then 3xy = 11 - 11² or xy <0 and either x or y must be negative and not both being positive as required.

If y – x = 11³ then 3xy = 1 - 11³ or xy <0 which is the same as the previous case.