## FE from USAMO 2002

For discussing Olympiad Level Algebra (and Inequality) problems
Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

### FE from USAMO 2002

Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x^2 - y^2) = x f(x) - y f(y)$ for all pairs of real numbers $x$ and $y$
This was freedom. Losing all hope was freedom.

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

### Re: FE from USAMO 2002

Let $P(x,y)$ denotes the assertion.
$P(0,0) \Rightarrow f(0)=0$
$P(x,0) \Rightarrow f(x^2)=xf(x)$
So, $f(a-b)=f(a)-f(b)$ when $a,b \in \mathbb{R^+}$
It can be proved that this statement is true for negatives too.
That is, $f(a-b)=f(a)-f(b)$ for all $a,b \in \mathbb{R}$

Now, $P(x+1,x) \Rightarrow f(2x+1)=(x+1)f(x+1) - xf(x) = f(x+1) + x(f(x+1)-f(x)) = f(x+1)+xf(1)$
$\Rightarrow f(2x+1)-f(x+1)=xf(1)$
$\Rightarrow f(x)=xf(1)$

So, all such functions are $f(x)=kx$
This was freedom. Losing all hope was freedom.

Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

### Re: FE from USAMO 2002

The fact that $f(x^2)=xf(x)$
implies that $f(x^2-y^2)=f(x^2)-f(y^2)$
so it implies that $f(a-b)=f(a)-f(b)$ for $a$ and $b$ positive perfect square; why do you say that this expression is true also for $a$ and $b$ positive numbers (and so no necessary positive perfect squares...)?
And why after do you say that is true also for $a$ and $b$ negative? (and so for all integers $a$ and $b$?) Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

### Re: FE from USAMO 2002

Katy729 wrote:
Mon Apr 23, 2018 11:31 am
The fact that $f(x^2)=xf(x)$
implies that $f(x^2-y^2)=f(x^2)-f(y^2)$
so it implies that $f(a-b)=f(a)-f(b)$ for $a$ and $b$ positive perfect square; why do you say that this expression is true also for $a$ and $b$ positive numbers (and so no necessary positive perfect squares...)?
And why after do you say that is true also for $a$ and $b$ negative? (and so for all integers $a$ and $b$?) You know, every positive real number is the square of another real number.

And, $P(0,x)$ implies $f(-x^2)=-xf(x)$ from which we can deduce that $f(-a)=-f(a)$. It gives us $f(a-b)=f(a)-f(b)$ for negative reals.
Hope you understand.
This was freedom. Losing all hope was freedom.

Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

### Re: FE from USAMO 2002

Ahh, I'm not sure. But I try:
You show that $f(-x^2)=-xf(x)$ but we know also that
$f(x^2)=xf(x)$ so $-f(-x^2)=f(x^2)$
and so $f(-x^2)=-f(x^2)$ and if we put $x^2=a$ we obtein
$f(-a)=-f(a)$ but we know that
$f(x^2-y^2)=xf(x)-yf(y)$ for positive real numbers; so using the fact that $f(-x^2)=-xf(x)$ to obtein that
$f(x^2-y^2)=-f(-x^2)+f(-y^2)$
so if we put $x^2=a$ and $y^2=b$ we obtein
$f(a-b)=-f(-a)+f(-b)$ but we know that $f(-a)=-f(a)$
so $f(a-b)=f(a)-f(b)$. But why $a$ and $b$ can be negative?? 