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# Understand the problem

Find all functions $f : \mathbb R \rightarrow \mathbb R$ such that $(x + y)(f(x) - f(y)) = (x -y)f(x + y)$ for all $x, y\in \mathbb R$

##### Source of the problem
Singapore Team Selection Test 2008
##### Topic
Functional Equations
Medium
##### Suggested Book
Functional Equations by BJ Venkatachala

Do you really need a hint? Try it first!

Play with choices of $x,y$ belonging to $\{0,1,-1\}$. Show that $f(0)=0$.

Play with choices of $x,y$ from the set $\left\{-\frac{z}{2},\frac{z}{2}+1,\frac{z}{2}-1\right\}$.

Following hint 2, you should be able to get equations that can be added to cancel some terms.

Putting $x=\frac{z}{2}-1,y=\frac{z}{2}+1$
we get $f(\frac{z}{2}-1)-f(\frac{z}{2}+1)=-\frac{2}{z}f(z)$ Putting $z=\frac{z}{2}+1,y=-\frac{z}{2}$ we get $f(\frac{z}{2}+1)-f(-\frac{z}{2})=(z+1)f(1)$ $x=\frac{z}{2},y=\frac{z}{2}-1$ gives $f(-\frac{z}{2})-f(\frac{z}{2}-1)=(z-1)f(-1)$

Adding, we get $f(z)=\frac{f(1)+f(-1)}{2}z^2+\frac{f(1)-f(-1)}{2}z$.

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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