jmc

Question

Let f(x) be a function defined for all positive real numbers satisfying the conditions f(x) > 0 for all x > 0 and f(x - y) = f(xy) + 2for all x > y > 0. Determine f(2009).

Accepted Answer

First, we claim there exist positive real numbers

x

and

y

so that

x - y = x y = 2009.

From these equations,

x^{2} - 2 x y + y^{2} = 200 9^{2},

so

x^{2} + 2 x y + y^{2} = 200 9^{2} + 4 \cdot 2009.

Then

x + y = 200 9^{2} + 4 \cdot 2009,

so by Vieta's formulas,

x

and

y

are the roots of

t^{2} - (200 9^{2} + 4 \cdot 2009) t + 2009 = 0.

(The discriminant of this quadratic is

200 9^{2},

so it does have real roots.)

Then for these values of

x

and

y,

f (2009) = f (2009) + 2 .

Let

a = f (2009),

so

a = a + 2 .

Squaring both sides, we get

a^{2} = a + 2,

so

a^{2} - a - 2 = 0.

This factors as

(a - 2) (a + 1) = 0.

Since

a

is positive,

a = 2 .