Define f1(x) = √(x2 + 48) and fn(x) = √(x2 + 6fn-1(x) ). Find all real solutions to fn(x) = 2x.
Solution
Answer: For each n, x = 4 is the only solution.
Obviously x = 4 is a solution. Since fn(x) >= 0, any solution must be non-negative. So we restrict attention to x ≥ 0.
Suppose x < 4. We show by induction that fn(x) > 2x. For n = 1, the claim is equivalent to 4x2 < x2 + 48, or x2 < 16, which is true. So suppose the result is true for n. Then x2 + 6fn(x) > x2 + 12x. But x < 4, so 3x2 < 12x, so 4x2 < x2 + 12x. Hence fn+1(x) > 2x, as required.
An exactly similar argument shows that fn(x) < 2x for x > 4. Hence x = 4 is the only solution.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002