# Functional equation with only two solutions?

Recently I started studying functional equations. Now I'm trying to find all solutions to the following functional equation:
$$(f(2x))^3=f(4x)((f(x))^2+xf(x)).$$
Unfortunately, I was able to show only few things about this equation, moreover, I wasn't even able to find $f(0)$ using substitution method. What I got (only by guessing) that there are at least two solutions: $f(x)=0$ and $f(x)=x$. But I don't know if there are any others. Any help would be very appreciated.

• If you assume that $f$ is differentiable, then differentiating the equation you get $6 f ( 2 x ) ^ 2 f ' ( 2 x ) = 4 f ' ( 4 x ) \big( f ( x ) ^ 2 + x f ( x ) \big) + f ( 4 x ) \big( 2 f ( x ) f ' ( x ) + f ( x ) + x f ' ( x ) \big)$, which by letting $x = 0$ gives $f ( 0 ) = 0$. – Mohsen Shahriari May 26 '17 at 16:18

## 1 Answer

Assuming that $f(x)$ has a power series expansion $\;f(x)=\sum_{n=0}^\infty c_n x^n\;$ then the functional equation rewritten as $\;0=f(4x)((f(x))^2+xf(x))-(f(2x))^3\;$ implies $\;0 = c_0^2 x +O(x)^2\;$ and so $\;c_0=0.\;$ Assuming $\;c_0=0\;$ the function eqution implies $\;0=4c_1^2(c_1-1)x^3 +O(x)^4\;$ and so $\;c_1=0\;$ or $\;c_1=1\;.$ In either case $\;c_n=0\;$ for all $\;n>1\;$ and so $f(x)=0$ or $f(x)=x$ for all $x$.