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Difference between revisions of "2003 Pan African MO Problems/Problem 6"

(Solution to Problem 6 -- epic functional equation problem)
 
m (→‎Solution)
 
Line 9: Line 9:
 
By letting <math>x = 0, y = n</math>, we have <math>f(0) - f(n^2) = n (f(0) - f(n))</math>, and by letting <math>x = 0, y = -n</math>, we have <math>f(0) - f(n^2) = -n (f(0) - f(-n))</math>.  Substitution and simplification results in
 
By letting <math>x = 0, y = n</math>, we have <math>f(0) - f(n^2) = n (f(0) - f(n))</math>, and by letting <math>x = 0, y = -n</math>, we have <math>f(0) - f(n^2) = -n (f(0) - f(-n))</math>.  Substitution and simplification results in
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
n (f(0) - f(n)) &= -n (f(0) - f(n)) \\
+
n (f(0) - f(n)) &= -n (f(0) - f(-n)) \\
 
f(0) - f(n) &= -f(0) + f(-n) \\
 
f(0) - f(n) &= -f(0) + f(-n) \\
 
2 f(0) &= f(n) + f(-n).
 
2 f(0) &= f(n) + f(-n).

Latest revision as of 13:01, 25 January 2020

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.

See Also

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