AIME 2008 II · 第 15 题

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 cubes 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 $2n = 362\implies n =\boxed{181}$ in the second condition works (we know $b_k$ is odd so we don't need to find $m$). So we're done.

Solution 3 (Big Bash)

Let us generate numbers $1$ to $1000$ for the second condition, for squares. We know for $N$ to be integer, the squares must be odd. So we generate $N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801$. $N$ cannot exceed $1000$ since it is AIME problem. Now take the first criterion, let $a$ be the smaller consecutive cube. We then get:

$N^2 = (A + 1)^3 - A^3$ $N^2 - 1 = 3A^2 + 3A$ $(N + 1)(N - 1) = 3A(A + 1)$

Now we know either $N + 1$ or $N - 1$ must be factor of $3$, hence $N = 1 \pmod 3$ or$N = 2 \pmod 3$. Only $1, 73, 181, 721$ satisfy this criterion. Testing each of the numbers in the condition yields $181$ as the largest that fits both, thus answer $= \boxed{181}$.

Video Solution

2008 AIME II #15

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