Difference between revisions of "2005 Alabama ARML TST Problems/Problem 14"

Revision as of 14:46, 1 March 2008

Problem

Find the fourth smallest possible value of $x+y$ where x and y are positive integers that satisfy the following equation:

$x^2-2y^2=1$ .

Solution

$x^2-2y^2=1$ means that $x$ is odd. We can let $x=2x_1-1$ :

$4x_1^2-4x_1-2y^2=0\Longrightarrow 2x_1^2-2x_1=y^2$

y is even, $y=2y_1$ .

$2x_1^2-2x_1=4y_1^2\Longrightarrow x_1^2-x_1=x_1(x_1-1)=2y^2$

We need to find all integers $x_1$ such that $x_1(x_1-1)$ is twice a perfect square.

Since $x_1$ and $x_1-1$ are relatively prime, then either $x_1$ is a perfect square or twice a perfect square, and the same for $x_1-1$ .

Let's say that $x_1$ is the perfect square. Then x_1 is odd. We check some:

$x_1=9$ works.

$x_1=289$ also works.

We should be able to find some smaller ones when $x_1$ is twice a perfect square, so we try that.

$x_1=2$ works.

$x_1=50$ works.

We need to make sure that there is no others below 289. We check all other twice perfect squares, so 289 is the smallest $x_1$ , and then $y_1=204$ . Then we get $x=577$ and $y=408$ . $x+y=\boxed{985}$ .

@@ Line 30: / Line 30: @@
 ==See also==
+{{ARML box|year=2005|state=Alabama|num-b=13|num-a=15}}
+[[Category:Intermediate Number Theory Problems]]

2005 Alabama ARML TST (Problems)
Preceded by: Problem 13	Followed by: Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Difference between revisions of "2005 Alabama ARML TST Problems/Problem 14"

Revision as of 14:46, 1 March 2008

Problem

Solution

See also