2008 AIME II Problems/Problem 15
Problem
Find the largest integer satisfying the following conditions:
- (i)
can be expressed as the difference of two consecutive cubes;
- (ii)
is a perfect square.
Solution
Solution 1
Write , or equivalently,
.
Since and
are both odd and their difference is
, they are relatively prime. But since their product is three times a square, one of them must be a square and the other three times a square. We cannot have
be three times a square, for then
would be a square congruent to
modulo
, which is impossible.
Thus is a square, say
. But
is also a square, say
. Then
. Since
and
have the same parity and their product is even, they are both even. To maximize
, it suffices to maximize
and check that this yields an integral value for
. This occurs when
and
, that is, when
and
. This yields
and
, so the answer is
.
Solution 2
Suppose that the consecutive cubes are and
. We can use completing the square and the first condition to get:
where
and
are non-negative integers. Now this is a Pell equation, with solutions in the form
. However,
is even and
is odd. It is easy to see that the parity of
and
switch each time (by induction). Hence all solutions to the first condition are in the form:
where
. So we can (with very little effort) obtain the following:
. It is an AIME problem so it is implicit that
, so
. It is easy to see that
is strictly increasing by induction. Checking
in the second condition works (we know
is odd so we don't need to find
). So we're done.
Solution 3 (Big Bash)
Let us generate numbers to
for the second condition, for squares. We know for
to be integer, the squares must be odd. So we generate
.
cannot exceed
since it is AIME problem. Now take the first criterion, let
be the smaller consecutive cube. We then get:
Now we know either or
must be factor of
, hence
. Only
satisfy this criterion. Testing each of the numbers in the condition yields
as the largest that fits both, thus answer
.
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last problem | |
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