[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.

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