Find all functions $f: \mathbb R_{>0} \rightarrow \mathbb R_{>0}$ such that $f(f(f(x)f(y)))=f(x)f(y^2)$ for all $x, y \in \mathbb R_{>0}$.

I made this problem myself.

  • 1
    $\begingroup$ f(x) = k gives an entire family of functions. $\endgroup$ Jul 22, 2016 at 20:02
  • 4
    $\begingroup$ @ChrisCudmore That doesn't work for every $k$. $\endgroup$
    – wythagoras
    Jul 22, 2016 at 20:03

3 Answers 3


$f(x)f(1)=f(f(f(x)f(1)))=f(f(f(1)f(x)))=f(1)f(x^2)$. Therefore $f(x)=f(x^2)$ for all $x$.

Now $f(x)^2=f(x)f(x^2)=f(f(f(x)f(x)))=f(f(f(x)))$. So $f(f(x))=f(f(x)^2)=f(f(f(f(x))))=f(f(x))^2$. Therefore $f(f(x))=1$ for all $x$. But now $f(x)^2=f(f(f(x)))=1$, so $f(x)=1$.

The only solution is the constant function $f(x)=1$.

  • 1
    $\begingroup$ That's doing my brain in... how do you get $f(f(f(f(x)))) = f(f(x))^2$? I just about get every other part though - nice solution! $\endgroup$
    – Shuri2060
    Jul 22, 2016 at 21:21
  • 1
    $\begingroup$ With $f(z)^2=f(f(f(z)))$, substitute $z=f(x)$. $\endgroup$
    – f''
    Jul 22, 2016 at 21:29
  • $\begingroup$ Thanks. Out of interest, is the solution to $f(x) = f(x^2)$ a constant in general (out of context of the question)? $\endgroup$
    – Shuri2060
    Jul 22, 2016 at 21:33
  • 1
    $\begingroup$ If $f$ must be continuous over $\mathbb{R}^+$, then yes, because $\lim_{x\rightarrow1}f(x)$ would not exist if it was not constant. $\endgroup$
    – f''
    Jul 22, 2016 at 21:41
  • 2
    $\begingroup$ most appropriate username for this solution $\endgroup$
    – elias
    Jul 22, 2016 at 22:03

Extremely-partial initial observations:

The LHS is symmetrical in x,y so the RHS is too so $f(x)f(y^2)=f(x^2)f(y)$ so $f(x^2)/f(x)$ is constant; say it equals $k$. Now our equation is $f(f(f(x)f(y)))=kf(x)f(y)$ or $f(f(z))=kz$ when $z=f(x)f(y)$.


morally the first of those observations says f looks like a logarithm and the second says it looks like a scaling, and the two together suggest that actually it'll have to be constant (hence identically 1).


Incomplete Answer:

$f$ cannot be a polynomial with degree larger than 0.


Let f(x) have degree $n$.

Let us consider the degrees of the $x$ and $y$ terms separately in each side of the equation.


The largest $x$ term has degree $n^3$.

The largest $y$ term has degree $n^3$.


The largest $x$ term has degree $n$.

The largest $y$ term has degree $2n$.

Hence it is impossible for $f$ to be a polynomial with a degree larger than 0, as we should be able to equate these degrees for both sides of the equation.

The only polynomial with degree 0 (of the form $f(x)=k$) which works is $f(x) = 1$ as it is the only solution to $f(x) = (f(x))^2$.$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ ($f(x) = 0$ doesn't work as $f(x) \not\in \mathbb{R}^+$)

  • $\begingroup$ Well, technically f(x) = 1 is a polynomial and it satisfies the equation given by wythagoras. $\endgroup$
    – C. Woods
    Jul 22, 2016 at 20:49
  • $\begingroup$ Yes, my comment was before you edited your answer. $\endgroup$
    – C. Woods
    Jul 22, 2016 at 20:50
  • $\begingroup$ Well maybe we commented/edited at the same time. I'll also point out that f(x) = 0 is not a solution because the question specifies that the function must map into the positive reals. $\endgroup$
    – C. Woods
    Jul 22, 2016 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.