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2008 AIME II Problems/Problem 15

Revision as of 14:32, 7 April 2008 by Azjps (talk | contribs) (solutions by (1) riliu and (2) Altheman)
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Problem

Find the largest integer $n$ satisfying the following conditions:

(i) $n^2$ can be expressed as the difference of two consecutive cubes;
(ii) $2n + 79$ is a perfect square.

Solution

Solution 1

Write $n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1$, or equivalently,

\[(2n + 1)(2n - 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2.\]

Since $2n + 1$ and $2n - 1$ are both odd and their difference is $2$, 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 $2n - 1$ be three times a square, for then $2n + 1$ would be a square congruent to $2$ modulo $3$, which is impossible.

Thus $2n - 1$ is a square, say $b^2$. But $2n + 79$ is also a square, say $a^2$. Then $(a + b)(a - b) = a^2 - b^2 = 80$. Since $a + b$ and $a - b$ have the same parity and their product is even, they are both even. To maximize $n$, it suffices to maximize $2b = (a + b) - (a - b)$ and check that this yields an integral value for $m$. This occurs when $a + b = 40$ and $a - b = 2$, that is, when $a = 21$ and $b = 19$. This yields $n = 181$ and $m = 104$, so the answer is $\boxed{181}$.


Solution 2

Suppose that the consecutive squares are $m$ and $m + 1$. We can use completing the square and the first condition to get: \[(2n)^2 - 3(2m + 1)^2 = 1\equiv a^2 - 3b^2\] where $a$ and $b$ are non-negative integers. Now this is a pell equation, with solutions in the form $(2 + \sqrt {3})^k = a_k + \sqrt {3}b_k,$ $k = 0,1,2,3,...$. However, $a$ is even and $b$ is odd. It is easy to see that the parity of $a$ and $b$ switch each time (by induction). Hence all solutions to the first condition are in the form: \[(2 + \sqrt {3})^{2k + 1} = a_k + \sqrt {3}b_k\] where $k = 0,1,2,..$. So we can (with very little effort) obtain the following: $(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)$. It is an AIME problem so it is implicit that $n < 1000$, so $2n < 2000$. It is easy to see that $a_n$ is strictly increasing by induction. Checking $2n = 362\implies n = 181$ in the second condition works (we know $b_k$ is odd so we don't need to find $m$). So we're done.

See also

2008 AIME II (Problems • Answer Key • Resources)
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