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Denote by $\mathbb{Q}^{+} $ the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ which satisfy the following equation for all $x,y \in \mathbb{Q}^{+}:$ $$f\left ( f(x)^{2}y \right )=x^{3}f(xy).$$

What I have tried is ... By substituting $y=1$, we get $$f\left ( f(x)^{2} \right )=x^{3}f(x).$$

Then, whenever $f(x)=f(y)$, we have $$x^{3}= \frac{f\left ( f(x)^{2} \right )}{f(x)}= \frac{f\left ( f(y)^{2} \right )}{f(y)}=y^{3}$$ which implies $x=y$, so the function $f$ is injective. and I think I'm stuck here, any help will be appreciated, thanks.

Note that it's an Olympiad question (IMO2010 SL, Problem A5).

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    $\begingroup$ If $z=f(1)$ then $f(z^2y)=f(y)$ so the injectivity gives $z^2y=y$ hence $f(1)=z=1$ (small attribute). $\endgroup$ – drhab Aug 13 '15 at 8:56
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    $\begingroup$ $f(x)=1/x$ fits the equation. $\endgroup$ – Arthur Aug 13 '15 at 9:02
  • $\begingroup$ @Arthur yes I think it does, but how? $\endgroup$ – Izhr Aug 13 '15 at 9:03
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    $\begingroup$ What do you mean "how"? It's just checking: $$\frac{1}{\frac{1}{x^2}y}=x^3\frac{1}{xy}$$Now, whether this is the only function that fits, I don't know. $\endgroup$ – Arthur Aug 13 '15 at 9:05
  • $\begingroup$ @Arthur you should prove It ..! $\endgroup$ – Izhr Aug 13 '15 at 9:06
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Ok, I'll complete with your solution...

$$f\left ( f(x)^{2}y \right )=x^{3}f(xy) \tag{1}.$$

$$f\left ( f(x)^{2} \right )=x^{3}f(x) \tag{2}.$$

Now replace $x$ by $xy$ in $(2)$ , now apply $(2)$ twice, second time to $\left (y, f(x)^{2} \right )$ instead of $(x,y)$ $$f\left ( f(xy)^{2} \right )=(xy)^{3}f(xy)=y^{3}f\left ( f(x)^{2}y \right )=f\left ( f(x)^{2} f(y)^{2} \right )$$ Since $f$ is injective, we get $$f(xy)^{2}=f(x)^{2}f(y)^{2}$$ $$f(xy)=f(x)f(y)$$

Therefore, $f$ is multiplicative. This also implies $f(1)=1$ and $f(x^n)=f(x)^n$ for all integers $n.$

Then the function equation $(1)$ can be re-written as $$f\left ( f(x) \right )^{2} f(y)=x^{3}f(x)f(y)$$ $$f\left ( f(x) \right )=\sqrt{x^{3}f(x)} \tag{3}$$

Let $g(x)=xf(x)$. Then, by $(3),$ we have $$g\left ( g(x) \right )=g\left (xf(x) \right )=xf(x)\cdot f\left (xf(x) \right )=xf(x)^{2} f\left ( f(x) \right )=xf(x)^{2}\sqrt{x^{3}f(x)}=\left ( xf(x) \right )^{5/2}=\left ( g(x) \right )^{5/2}$$

and, by induction, $$\underset{n+1}{\underbrace{g ( g(....g}}(x)....) )=\left ( g(x) \right )^{(5/2)^{n}} \tag{4}$$

for every positive integer $n.$

Consider $(4)$ for a fixed $x.$ The left-hand side is always rational, so $\left ( g(x) \right )^{(5/2)^{n}}$ must be rational for every $n$. We show that this is possible only if $g(x)=1$. Suppose that $g(x)\neq 1$ and let the prime factorization of $g(x)$ be $g(x)=p_{1}^{\alpha _{1}}... p_{k}^{\alpha _{k}}$ where $p_{1}...p_{k}$ are distinct primes and $\alpha _{1}...\alpha _{k}$ are nonzero integers. Then the unique prime factorization of $(4)$ is

$$\underset{n+1}{\underbrace{g ( g(....g}}(x)....) )=\left ( g(x) \right )^{(5/2)^{n}} =p_{1}^{(5/2)^{n}{\alpha _{1}}}...p_{k}^{(5/2)^{n}{\alpha _{k}}}$$

where the exponents should be integers. But this is not true for large values of $n$, for example $(\frac{5}{2})^{n} \alpha _{1}$ cannot be a integer number when $2^{n}\not{\mid } \alpha $. Therefore $g(x) \neq 1$ is impossible.

Hence, $g(x)=1$ and thus $f(x)=\frac{1}{x}$ for all $x$.

The function $f(x)=\frac{1}{x}$ satisfies the equation $(1)$:

$$f\left ( f(x)^{2}y \right )=\frac{1}{f(x)^{2}y}=\frac{1}{(\frac{1}{x}^{2})y}=\frac{x^{3}}{xy}=x^{3}f(x).$$

AND WE'RE DONE. "SP3ED"

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    $\begingroup$ Oh, genies, wonderful solution, much thanks $\endgroup$ – Izhr Aug 13 '15 at 11:07
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    $\begingroup$ This is (beginning from second line) word by word official solution from IMO 2010 shortlist problem A5 found here: imo-official.org/problems/IMO2010SL.pdf . It is a good practice to cite sources, especially if you just copy the content. $\endgroup$ – Sil Apr 7 '18 at 7:35
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Let $P(x,y)$ be the assertion: $$ f(f(x)^2y)=x^3f(xy)\space\forall x,y\in\mathbb{Q^+} $$ We get (due to the injectivity): $$ P(1,y): f(f(1)^2y)=f(y)\iff f(1)^2y=y\iff f(1)=1 $$ $$ P(x,f(y)^2): f(f(x)^2f(y)^2)=x^3f(xf(y)^2)=x^3y^3f(xy) $$ $$ P(xy,1): f(f(xy)^2)=x^3y^3f(xy) $$ If we combine the last to results, we get: $$ f(f(xy)^2)=f(f(x)^2f(y)^2)\iff f(xy)=f(x)f(y) $$ Now define $g(x):=xf(x)$: $$ P(x,x^2): f(x^2f(x)^2)=x^3f(x^3)\iff f(x)^2f(f(x))^2=x^3f(x^3)\iff g(f(x))=g(x)^{\frac{3}{2}} $$ If we iterate this relation, we obtain: $$ g(x)^{\frac{3^n}{2^n}}=g(f^n(x)) $$ Where $f^n$ denotes the $n$-th iteration of $f$. Therefore, since $g(x)^{\frac{3^n}{2^n}}$ is rational, $g(x)^{\frac{1}{2^n}}$ needs to be rational to for all $n\in\mathbb{N_0}$ and thus $g(x)=1$. Therfore, we obtain the only solution: $$ f(x)=\frac{1}{x}\space\forall x\in\mathbb{Q^+} $$

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$f(f(x)^2f(z)^2)=x^3f(xf(z)^2)=x^3z^3f(xz)$ so $f(x)f(z)=f(xz)f(1)=f(xz)$ because $f$ is injective. So $f(x)$ is defined by all the values of $f(p)$ for $p$ prime.
$f(f(x))^2=x^3f(x)$ so $f(p)=g(p)^2/p$ for a rational number $g(p)$.
$$f(f(p))=f(g(p)^2/p)^2=p^3g(p)^2/p=p^2g(p)^2\\ f(g(p)^2/p)=pg(p)\\ f(g(p))^2=(g(p))^3$$ So $g(p)$ is a square number, $g(p)=h(p)^2$. Then $f(h(p))^2=(h(p))^3$, and for the same reason, $h(p)=i(p)^2=j(p)^4=\cdots$
So $g(p)=1$ for all $p$, and $f(x)=1/x$

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