## FE from USAMO 2002

- Atonu Roy Chowdhury
**Posts:**63**Joined:**Fri Aug 05, 2016 7:57 pm**Location:**Chittagong, Bangladesh

### 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**Location:**Chittagong, Bangladesh

### 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 $f(x)=kx$

$P(0,0) \Rightarrow f(0)=0$

$P(x,0) \Rightarrow f(x^2)=xf(x)$

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.

### 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 $b$?)

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**Location:**Chittagong, Bangladesh

### Re: FE from USAMO 2002

You know, every positive real number is the square of another real number.Katy729 wrote: ↑Mon Apr 23, 2018 11:31 amThe 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 $b$?)

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.

### 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??

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??