Problem 47: APMO 2002, Problem 5

by henderson, Sep 4, 2016, 10:48 AM

$\color{red}\bf{Problem \ 47}$ Let $\bf\mathbb{R}$ denote the set of all real numbers. Find all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying:
$\text{(i)}$ there are only finitely many $s$ in $\bf\mathbb{R}$ such that $f(s)=0;$
$\text{(ii)}$ $f(x^4+y)=x^3f(x)+f(f(y))$ for all $x,y\in\mathbb{R}.$
$($ My solution

$\color{red}\bf{My \ solution}$ Let $P(x,y)$ be the assertion.

$\boxed{\color{blue}\textbf{1}}$ $\quad$ $P(1,0)$ $\Longrightarrow$ $f(f(0))=0.$ $\quad P(0,0)$ $\Longrightarrow$ $f(0)=0.$

$\boxed{\color{blue}\textbf{2}}$ $\quad$ $P(0,y)$ $\Longrightarrow$ $f(y)=f(f(y)).$

$\boxed{\color{blue}\textbf{3}}$ $\quad$ $P(x,0)$ $\Longrightarrow$ $f(x^4)=x^3f(x).$

$\boxed{\color{blue}\textbf{4}}$ $\quad$ $P(-x,y)$ $\Longrightarrow$ $f(x^4+y)=-x^3f(-x)+f(f(y)).$ But it's given that $f(x^4+y)=x^3f(x)+f(f(y)),$ so $x^3f(x)=-x^3f(-x)$ $\Longrightarrow$ $f(x)=-f(-x)$ for any nonzero real number $x.$ Since $f(0)=0=-f(0),$ $f(x)=-f(-x)$ holds for all $x\in\mathbb{R}.$ In other words, $f$ is odd.

$\boxed{\color{blue}\textbf{5}}$ $\quad$ Let $f(s)=0,$ where $s\not= 0.$ Then, using $\boxed{\color{blue}\textbf{3}},$ we obtain $f(s^4)=0.$ Repeating this process, we conclude that $f(s^{4^k})=0$ for any nonnegative integer $k.$ But, then there will be infinitely many $s$ satisfying $f(s)=0,$ which is contradiction to $\text{(i)}.$ Hence, $f(s)=0$ $\iff$ $s=0.$

$\boxed{\color{blue}\textbf{6}}$ $\quad$ Substituting $x^3f(x)$ and $f(f(y))$ with $f(x^4)$ and $f(y),$ respectively, in the original equation, we get
$f(x^4+y)=f(x^4)+f(y).$ Since $x^4$ can get any nonnegative real value, $f(x+y)=f(x)+f(y)$ holds for all $y\in \mathbb{R}$ and any nonnegative real number $x.$
Using the fact that $f$ is odd, we have
$-f(-x-y)=f(x+y)=f(x)+f(y)=-f(-x)-f(-y) \Longrightarrow f(-x-y)=f(-x)+f(-y)$ for all $y\in \mathbb{R}$ and any nonnegative real number $x$ $\Longrightarrow$
$f(x+y)=f(x)+f(y)$ holds for all $y\in \mathbb{R}$ and any nonpositive real number $x.$ Therefore,
$f(x+y)=f(x)+f(y)$ holds for all $x,y\in \mathbb{R}.$

$\boxed{\color{blue}\textbf{7}}$ $\quad$ Let $f(a)=f(b)$ and $a=b+c.$ Also, let $Q(x,y)$ be the assertion $f(x+y)=f(x)+f(y).$
Then, $Q(b,c)$ gives $f(c)=0.$ But, as shown in $\boxed{\color{blue}\textbf{5}},$ $f(s)=0$ $\iff$ $s=0.$ Hence, $c=0$ $\Longrightarrow$ $a=b,$ which means that $f$ is injective.

$\boxed{\color{blue}\textbf{8}}$ $\quad$ Combining $\boxed{\color{blue}\textbf{2}}$ and $\boxed{\color{blue}\textbf{7}},$ we obtain that $f(x)=x$ for all $x\in \mathbb{R}.$ Obviously, it satisfies the problem conditions.