2003 Pan African MO Problems/Problem 6

Problem

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that: $f(x^2)-f(y^2)=(x+y)(f(x)-f(y))$ for $x,y \in \mathbb{R}$ .

Solution

By letting $x = 0, y = n$ , we have $f(0) - f(n^2) = n (f(0) - f(n))$ , and by letting $x = 0, y = -n$ , we have $f(0) - f(n^2) = -n (f(0) - f(-n))$ . Substitution and simplification results in $\begin{align*} n (f(0) - f(n)) &= -n (f(0) - f(-n)) \\ f(0) - f(n) &= -f(0) + f(-n) \\ 2 f(0) &= f(n) + f(-n). \end{align*}$ Therefore, if $f(n) = f(0) + m$ , where $m$ is a real number, then $f(-n) = f(0) - m$ . Thus, the function $f(x)$ has rotational symmetry about $(0,f(0))$ .

Additionally, multiplying both sides by $x - y$ results in $(x-y)(f(x^2)-f(y^2))=(x^2 - y^2)(f(x)-f(y))$ , and rearranging results in $\frac{f(x^2) - f(y^2)}{x^2 - y^2} = \frac{f(x)-f(y)}{x-y}$ . The equation seems to resemble a rearranged slope formula, and the rotational symmetry seems to hint that $f(x)$ is a linear function.

To prove that $f(x)$ must be a linear function, we need to prove that the slope is the same from $(0,f(0))$ to all the other points of the function. By letting $x = n, y = 0$ , we have $\frac{f(n^2) - f(0)}{n^2} = \frac{f(n) - f(0)}{n}$ . Additionally, by letting $x = n, y = 1$ , we have $\frac{f(n^2) - f(1)}{n^2-1} = \frac{f(n)-f(1)}{n-1}$ . By rearranging the prior equation, we have $\begin{align*} f(n^2) - f(1) &= (n^2 - 1) \cdot \frac{f(n)-f(1)}{n-1} \\ f(n^2) - f(1) &= (n+1) f(n) - (n+1) f(1) \\ f(n^2) &= (n+1) f(n) - n f(1) \\ f(1) &= \frac{(n+1) f(n) - f(n^2)}{n} \end{align*}$ The slope from points $(0,f(0))$ and $(1,f(1))$ is $\frac{f(1)-f(0)}{1-0} = f(1) - f(0)$ . By substitution, $\begin{align*} f(1) - f(0) &= \frac{(n+1) f(n) - f(n^2)}{n} - \frac{n f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(n^2) - n f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(n^2) + f(0) - f(0) - n f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(0) - n f(0)}{n} - \frac{f(n^2) - f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(0) - n f(0)}{n} - n \cdot \frac{f(n^2) - f(0)}{n^2} \\ &= \frac{n f(n) + f(n) - f(0) - n f(0)}{n} - n \cdot \frac{f(n) - f(0)}{n} \\ &= \frac{n f(n) + f(n) - f(0) - n f(0)}{n} - \frac{n f(n) - n f(0)}{n} \\ &= \frac{f(n) - f(0)}{n}. \end{align*}$ The slope from point $(0,f(0))$ to $(1,f(1))$ is the same as the slope from point $(0,f(0))$ to any other point on the function, so $f(x)$ must be a linear function.

Let $f(x) = mx+b$ . Using the function on the original equation results in $\begin{align*} f(x^2)-f(y^2) &= (x+y)(f(x)-f(y)) \\ mx^2 + b - my^2 - b &= (x+y)(mx+b-my-b) \\ mx^2 - my^2 &= (x+y)(mx-my) \\ mx^2 - my^2 &= mx^2 - my^2 \end{align*}$ Thus, $f(x)$ can be any linear function, so $\boxed{f(x) = mx+b}$ , where $m, b$ are real numbers.

2003 Pan African MO (Problems)
Preceded by Problem 5	1 • 2 • 3 • 4 • 5 • 6	Followed by Last Problem
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2003 Pan African MO Problems/Problem 6

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