Difference between revisions of "2008 AIME II Problems/Problem 15"

Revision as of 18:55, 22 May 2017

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 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.

@@ Line 22: / Line 22: @@
 (2 + \sqrt {3})^{2k + 1} = a_k + \sqrt {3}b_k
 </cmath>
-where <math>k = 0,1,2,..</math>. So we can (with very little effort) obtain the following: <math>(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)</math>. It is an AIME problem so it is implicit that <math>n < 1000</math>, so <math>2n < 2000</math>. It is easy to see that <math>a_n</math> is strictly increasing by induction. Checking <math>2n = 362\implies n = 181</math> in the second condition works (we know <math>b_k</math> is odd so we don't need to find <math>m</math>). So we're done.
+where <math>k = 0,1,2,..</math>. So we can (with very little effort) obtain the following: <math>(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)</math>. It is an AIME problem so it is implicit that <math>n < 1000</math>, so <math>2n < 2000</math>. It is easy to see that <math>a_n</math> is strictly increasing by induction. Checking <math>2n = 362\implies n =boxed{181}</math> in the second condition works (we know <math>b_k</math> is odd so we don't need to find <math>m</math>). So we're done.
 == See also ==

2008 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Last problem
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