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.