Mock AIME I 2015 Problems/Problem 9
Revision as of 23:25, 10 January 2019 by Blizzardwizard (talk | contribs)
Since is a multiple of
, let
.
We can rewrite the first and second conditions as:
(a)
is a perfect square, or
is a perfect square.
(b)
is a power of
, so it follows that
,
, and
are all powers of
.
Now we use casework on
. Since
is a power of
,
is
or
or
.
If
, then no value of
makes
.
If
or
, then no value of
that is a power of
makes
a perfect square.
If
, then
and
for
solutions.
If
, then
and
for
solutions.
If
, then
and
for
solutions.
If
, then
and
for
solutions.
This is a total of
solutions.