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2005 Alabama ARML TST Problems/Problem 14

Revision as of 14:13, 5 May 2009 by 5849206328x (talk | contribs) (Added second solution)

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

Solution 1

$x^2-2y^2=1$ 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 $x+y=\boxed{985}$.

Solution 2

We quickly find the first solution, $(x,y)=(3,2)$. Factoring, we get \[(3-2\sqrt{2})(3+2\sqrt{2})=1\] We can square both sides to get \[(3-2\sqrt{2})^2(3+2\sqrt{2})^2=1^2 \Rightarrow (17-12\sqrt{2})(17+12\sqrt{2})=1\] So $(x,y)=(17,12)$ is another solution.

This gives us a way to generate whatever solutions we want to the equation. Raising the first equation to the fourth power gives us \[(577-408\sqrt{3})(577+408\sqrt{3})=1\] The answer is $577+408=\boxed{985}$.

See also

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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