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 squares 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.
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
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