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.