Maths Olympiad Prep

Track / Stage 5 / 65 of 400 #665 of 1964

Problem 665

AIME late
Algebra Difficulty 5.1 Prove it Singapore Mathematical Olympiad (SMO) · Singapore

Find all functions f:RRf: \mathbb{R} \to \mathbb{R} so that
(x+y)(f(x)f(y))=(xy)f(x+y) (x+y)(f(x)-f(y)) = (x-y)f(x+y)
for all x,yRx, y \in \mathbb{R}.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Suppose that ff is a solution. Let
a=12(f(1)f(1)),b=12(f(1)+f(1)) a = \frac{1}{2}(f(1) - f(-1)), \quad b = \frac{1}{2}(f(1) + f(-1))
and g(x)=f(x)axbx2g(x) = f(x) - a x - b x^2. Then
(x+y)(g(x)g(y))=(xy)g(x+y) (x+y)(g(x)-g(y)) = (x-y)g(x+y)
and g(1)=g(1)=0g(1) = g(-1) = 0. Letting y=1y = 1 and y=1y = -1 above give
(x+1)g(x)=(x1)g(x+1)xg(x+1)=(x+2)g(x). \begin{aligned} (x+1)g(x) &= (x-1)g(x+1) \\ xg(x+1) &= (x+2)g(x). \end{aligned}
Thus
x(x+1)g(x)=x(x1)g(x+1)=(x1)(x+2)g(x) x(x+1)g(x) = x(x-1)g(x+1) = (x-1)(x+2)g(x)
for all xx. So g(x)=0g(x) = 0 for all xx. Hence f(x)=ax+bx2f(x) = a x + b x^2. We can check directly that any function of this form (for some a,bRa, b \in \mathbb{R}) satisfies the given equation.

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