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

(solutions by (1) riliu and (2) Altheman)
 
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
−
Write <math>n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1</math>, or equivalently,
+
Write <math>n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1</math>, or equivalently, <math>(2n + 1)(2n - 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2</math>.
−
 
 
−
<cmath>(2n + 1)(2n - 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2.</cmath>
 
  
 
Since <math>2n + 1</math> and <math>2n - 1</math> are both odd and their difference is <math>2</math>, 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 <math>2n - 1</math> be three times a square, for then <math>2n + 1</math> would be a square congruent to <math>2</math> modulo <math>3</math>, which is impossible.
 
Since <math>2n + 1</math> and <math>2n - 1</math> are both odd and their difference is <math>2</math>, 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 <math>2n - 1</math> be three times a square, for then <math>2n + 1</math> would be a square congruent to <math>2</math> modulo <math>3</math>, which is impossible.
Line 15: Line 13:
 
Thus <math>2n - 1</math> is a square, say <math>b^2</math>. But <math>2n + 79</math> is also a square, say <math>a^2</math>. Then <math>(a + b)(a - b) = a^2 - b^2 = 80</math>. Since <math>a + b</math> and <math>a - b</math> have the same parity and their product is even, they are both even. To maximize <math>n</math>, it suffices to maximize <math>2b = (a + b) - (a - b)</math> and check that this yields an integral value for <math>m</math>. This occurs when <math>a + b = 40</math> and <math>a - b = 2</math>, that is, when <math>a = 21</math> and <math>b = 19</math>. This yields <math>n = 181</math> and <math>m = 104</math>, so the answer is <math>\boxed{181}</math>.
 
Thus <math>2n - 1</math> is a square, say <math>b^2</math>. But <math>2n + 79</math> is also a square, say <math>a^2</math>. Then <math>(a + b)(a - b) = a^2 - b^2 = 80</math>. Since <math>a + b</math> and <math>a - b</math> have the same parity and their product is even, they are both even. To maximize <math>n</math>, it suffices to maximize <math>2b = (a + b) - (a - b)</math> and check that this yields an integral value for <math>m</math>. This occurs when <math>a + b = 40</math> and <math>a - b = 2</math>, that is, when <math>a = 21</math> and <math>b = 19</math>. This yields <math>n = 181</math> and <math>m = 104</math>, so the answer is <math>\boxed{181}</math>.
  
−
__TOC__
 
 
=== Solution 2 ===
 
=== Solution 2 ===
 
Suppose that the consecutive squares are <math>m</math> and <math>m + 1</math>. We can use completing the square and the first condition to get:
 
Suppose that the consecutive squares are <math>m</math> and <math>m + 1</math>. We can use completing the square and the first condition to get:

Revision as of 20:01, 17 April 2008

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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Problem 14 		Followed by
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