2005 Alabama ARML TST Problems/Problem 14

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 $x=2x_1-1$ for some $x_1>0$ :

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

y is even, $y=2y_1$ for some $y_1>0$ .

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

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, one of them is a perfect square and the other is twice a perfect square. Moreover, the perfect square must be odd.

We will now find four smallest solutions for $x_1$ . Obviously, these will give the four smallest solutions for $x+y$ .

Each time we examine whether the value $y_1=\sqrt{\frac{x_1(x_1-1)}2}$ is a positive integer.

$x_1=1$ gives $y_1=0$ which is not positive.
$x_1-1=1$ gives $y_1=1$ , hence $(x,y)=(3,2)$ .
$x_1=9$ gives $y_1=6$ , hence $(x,y)=(17,12)$ .
$x_1-1=9$ gives $y_1=\sqrt{9\cdot 5}$ .
$x_1=25$ gives $y_1=\sqrt{25\cdot 12}$ .
$x_1-1=25$ gives $y_1=\sqrt{25\cdot 13}$ .
$x_1=49$ gives $y_1=\sqrt{49\cdot 24}$ .
$x_1-1=49$ gives $y_1=\sqrt{49\cdot 25}=35$ , hence $(x,y)=(99,70)$ .
$x_1=81$ gives $y_1=\sqrt{81\cdot 40}$ .
$x_1-1=81$ gives $y_1=\sqrt{81\cdot 41}$ .
$x_1=121$ gives $y_1=\sqrt{121\cdot 60}$ .
$x_1-1=121$ gives $y_1=\sqrt{121\cdot 61}$ .
$x_1=169$ gives $y_1=\sqrt{169\cdot 84}$ .
$x_1-1=169$ gives $y_1=\sqrt{169\cdot 85}$ .
$x_1=225$ gives $y_1=\sqrt{225\cdot 112}$ .
$x_1-1=225$ gives $y_1=\sqrt{225\cdot 113}$ .
$x_1=289$ gives $y_1=\sqrt{289\cdot 144}=17\cdot 12 = 204$ , hence $(x,y)=(577,408)$ , and the answer is $x+y=\boxed{985}$ .

2005 Alabama ARML TST (Problems)
Preceded by: Problem 13	Followed by: Problem 15
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2005 Alabama ARML TST Problems/Problem 14

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