2008 AIME II Problems/Problem 15
Find the largest integer satisfying the following conditions:
- (i) can be expressed as the difference of two consecutive cubes;
- (ii) is a perfect square.
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 .
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 BAYUS)
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 or. Only satisfy this criterion. Testing each of the numbers in the condition yields as the largest that fits both, thus answer .
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