**2014 SMO (Senior-Rd1) 23.** Let be a positive integer, and let and . It is given that . Find the value of .

The first observation I made was that

.

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

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

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

Looking at the expression for in terms of , it looked like a complicated fraction. It seemed like a good idea to simplify it, since 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:**

We can do something similar for :

Remembering that , we can use to get

Since is positive, we must have . The answer is **5**.