Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 BMO Problems/Problem 2"

 
m (official wording)
 
Line 6: Line 6:
 
<math>
 
<math>
 
\displaystyle f(f(x) + y) = f(f(x) - y) + 4f(x)y
 
\displaystyle f(f(x) + y) = f(f(x) - y) + 4f(x)y
−
</math>.
+
</math>, for any <math> x,y \in \mathbb{R} </math>.
 
</center>
 
</center>
  

Latest revision as of 22:46, 4 May 2007

Problem

(Bulgaria) Find all functions $\displaystyle f : \mathbb{R} \mapsto \mathbb{R}$ such that

$\displaystyle f(f(x) + y) = f(f(x) - y) + 4f(x)y$, for any $x,y \in \mathbb{R}$.

Solution

We first note that $\displaystyle f(x) = 0$ is a solution to the equation. Henceforth we shall consider other solutions to the equations, i.e., functions $\displaystyle f$ such that for some $\displaystyle c$, $\displaystyle f(c) \neq 0$.

Setting $\displaystyle y = f(x)$ gives us $\displaystyle f(2f(x)) = f(0) + [f(x)]^2$.

We note that for any $\displaystyle x$, $\displaystyle f \left[ f(c) + \frac{x}{4f(c)} \right] - f \left[ f(c) - \frac{x}{4f(c)} \right] = f(c)\cdot \frac{x}{4f(c)} = x$, i.e., as $\displaystyle a$ and $\displaystyle b$ assume all real values, $\displaystyle f(b) - f(a)$ assume all real values.

Now, setting $\displaystyle x=b$ and $\displaystyle y = 2f(a) - f(b)$, we obtain

$\displaystyle f[2f(a)] = f\{ 2[f(b)-f(a)]\} + 4f(b)[2f(a)-f(b)]$,

or

$\displaystyle f\{ 2[f(b) - f(a)]\} = 4[f(a)]^2 +f(0) - 8f(a)f(b) + 4[f(b)]^2 + f(0) = \{ 2[f(b) - f(a)] \}^2 + f(0)$.

Since $\displaystyle 2[ f(b) - f(a) ]$ takes on all real values, it follows that for all $\displaystyle x$, $\displaystyle f(x) = x^2 + f(0)$. It is easy to see that any value of $\displaystyle f(0)$ will satisfy the desired condition. Thus the only solutions to the functional equation are $\displaystyle f(x) = 0$ and $\displaystyle f(x) = x^2 + c$, $\displaystyle c$ an arbitrary constant.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_BMO_Problems/Problem_2&oldid=14823"