## [Soln] 2014 Singapore MO (Senior Rd 1) Problem 23

2014 SMO (Senior-Rd1) 23. Let $n$ be a positive integer, and let $x = \frac{\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} + \sqrt{n}}$ and $y = \frac{\sqrt{n+2} + \sqrt{n}}{\sqrt{n+2} - \sqrt{n}}$. It is given that $14x^2 + 26xy + 14y^2 = 2014$. Find the value of $n$.

The first observation I made was that

$y = \displaystyle\frac{1}{x} \hspace{1em} -(1)$.

Certainly not a coincidence, sure to come in handy. The next thing I did was to substitute it into the given equation in $x$ and $y$ to make it into an equation in one variable:

\begin{aligned} 14x^2 + 26 + \displaystyle\frac{14}{x^2} &= 2014, \\ 14x^2 + \displaystyle\frac{14}{x^2} &= 1988, \\ x^2 + \frac{1}{x^2} &= 142 \hspace{1em} -(2). \end{aligned}

This looked very promising! A simple equation in $x$. I could convert it into a quadratic equation in $x^2$ and solve for it:

\begin{aligned} x^4 - 142x^2 + 1 &= 0, \\ x^2 &= \frac{142 \pm \sqrt{142^2 - 4(1)(1)}}{2}. \end{aligned}

Hmm… this line of thought didn’t seem so appealing anymore. That square root was not going to simplify, so the expression for $x$ would have a square root in a square root. Then I would have to substitute this into the expression for $x$ and try to solve for $n$ there… So $(2)$ seemed like an important piece of information, but maybe solving directly for $x$ was not the way to go.

Looking at the expression for $x$ in terms of $n$, it looked like a complicated fraction. It seemed like a good idea to simplify it, since $n$ was what I was going for anyway. A standard way to simplify such fractions is to multiply the top and bottom of the fraction by the denominator’s conjugate:

\begin{aligned} x & = \frac{(\sqrt{n+2} - \sqrt{n})^2}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+2}-\sqrt{n})} \\ &= \frac{n+2 - 2 \sqrt{n(n+2)} + n}{(n+2) - n} \\ &= n + 1 - \sqrt{n(n+2)}, \\ x^2 &= (n+1)^2 - 2(n+1)\sqrt{n(n+2)} + n(n+2). \end{aligned}

We can do something similar for $y$:

\begin{aligned} y & = \frac{(\sqrt{n+2} + \sqrt{n})^2}{(\sqrt{n+2}-\sqrt{n})(\sqrt{n+2}+\sqrt{n})} \\ &= \frac{n+2 + 2 \sqrt{n(n+2)} + n}{(n+2) - n} \\ &= n + 1 + \sqrt{n(n+2)}, \\ y^2 &= (n+1)^2 + 2(n+1)\sqrt{n(n+2)} + n(n+2). \end{aligned}

Remembering that $y = \frac{1}{x}$, we can use $(2)$ to get

\begin{aligned} 142 &= x^2 + y^2, \\ 142 &= [(n+1)^2 - 2(n+1)\sqrt{n(n+2)} + n(n+2)] \\ &\hspace{1em}+ [(n+1)^2 + 2(n+1)\sqrt{n(n+2)} + n(n+2)], \\ 142 &= 2(n+1)^2 + 2n(n+2), \\ 142 &= 2n^2 + 4n + 2 + 2n^2 + 4n, \\ 4n^2 + 8n + 2 &= 142, \\ n^2 + 2n - 35 &= 0, \\ (n+7)(n-5) &= 0.\end{aligned}

Since $n$ is positive, we must have $n = 5$. The answer is 5.