1998 JBMO Problems/Problem 3
Revision as of 01:34, 18 September 2020 by Duck master (talk | contribs) (fixed stupid arithmetic mistake. the second solution to the eq is ok now)
Find all pairs of positive integers such that
Solution
Note that is at least one. Then
is at least one, so
.
Write , where
. (We know that
is nonnegative because
.) Then our equation becomes
. Taking logarithms base
and dividing through by
, we obtain
.
Since divides the RHS of this equation, it must divide the LHS. Since
by assumption, we must have
, so that the equation reduces to
, or
. This equation has only the solutions
and
.
Therefore, our only solutions are , and
, and we are done.
See also
1998 JBMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All JBMO Problems and Solutions |