2005 Alabama ARML TST Problems/Problem 14

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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 $y=408$. $x+y=\boxed{985}$.

See also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 13
Followed by:
Problem 15
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