$f(x^2) = xf(x)$ implies that $ f(x) = mx$? Suppose a function $f : \mathbb{R} \to \mathbb{R} $ satisfies the relation $$f(x^2) = xf(x) \ \ \forall x$$  Does this imply $f$ must be a straight line, $f(x) = mx$? If so, why?  If not, are there other such functions?
 A: Let $f(x)=x$ if $x$ is an algebraic number, and $f(x)=0$ if $x$ is transcendental. 
A: Of course not.
To find the general solution of $f(x^2)=xf(x)$ :
First, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2308.pdf
Let $x_1=\ln x$ , $f_1(x_1)=f(x)$ ,
Then $f_1(2x_1)=e^{x_1}f_1(x_1)$
Then, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2303.pdf
Let $x_2=\ln x_1$ , $f_2(x_2)=f_1(x_1)$ ,
Then $f_2(x_2+\ln2)=e^{e^{x_2}}f_2(x_2)$
$x_2\to x_2\ln2$ :
$f_2(x_2\ln2+\ln2)=e^{e^{x_2\ln2}}f_2(x_2\ln2)$
$f_2((x_2+1)\ln2)=e^{2^{x_2}}f_2(x_2\ln2)$
$f_2(x_2\ln2)=\Theta_1(x_2)\prod_{x_2}e^{2^{x_2}}$, where $\Theta_1(x_2)$ is an arbitrary periodic function with unit period
According to http://en.wikipedia.org/wiki/Indefinite_product#Rules ,
$f_2(x_2\ln2)=\Theta_1(x_2)e^{\sum_{x_2}2^{x_2}}$, where $\Theta_1(x_2)$ is an arbitrary periodic function with unit period
According to http://en.wikipedia.org/wiki/Indefinite_sum#Antidifferences_of_exponential_functions ,
$f_2(x_2\ln2)=\Theta_1(x_2)e^{2^{x_2}}$, where $\Theta_1(x_2)$ is an arbitrary periodic function with unit period
$f_2(x_2\ln2)=\Theta_1(x_2)e^{e^{x_2\ln2}}$, where $\Theta_1(x_2)$ is an arbitrary periodic function with unit period
$f_2(x_2)=\Theta(x_2)e^{e^{x_2}}$, where $\Theta(x_2)$ is an arbitrary periodic function with period $\ln2$
$f(x)=\Theta(\ln\ln x)x$, where $\Theta(x)$ is an arbitrary periodic function with period $\ln2$
A: No, there are other such functions.  For example, define $f\colon\mathbb{R}\to\mathbb{R}$ by
$$
f(x) \;=\; \begin{cases} 2x & \text{if }x\text{ is algebraic,} \\[6pt] 3x & \text{if }x\text{ is transcendental.}\end{cases}
$$
Both algebraic and transcendental numbers are closed under the operation of squaring, and therefore $f(x^2) = x\,f(x)$ for all $x$.
