# 1982 IMO Problems/Problem 4

Prove that if is a positive integer such that the equation has a solution in integers , then it has at least three such solutions. Show that the equation has no solutions in integers for .

Suppose the equation has solution in integers with . Then, completing the cube yields . Using the substitution yields . Notice that equality directly implies that is also a solution to the original equation. Applying the transformation again yields that is also a solution. We show that these three solutions are indeed distinct: If then which only has solution which implies that is not a positive integer, a contradiction. Similarly, since the transformation from to and to is the same as the transformation from to , we have that the three solutions are pairwise distinct.

For the case , notice that . Considering all solutions modulo of the equation yields only . But, this implies that divides which is clearly not true.