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Find all $f:\mathbb{Q^+}\rightarrow \mathbb{Q^+}$ so that $\forall x,y\in\mathbb{Q^+}$ : $f(f(x)^2y)=x^3f(xy)$

so $f(0)=0$ when $x=y=0$

$f(f(1)^2)=f(1)$ when $x=y=1$

$f(f(x)^2)=x^3f(x)$ when $y=1$ and therefore $f$ is injective so $f(1)=1$

1 Answers1

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$x=xy,y=1\Longrightarrow f(f(xy)^2)=x^3y^3f(xy)$

$y=f(y)^2\Longrightarrow f(f(x)^2f(y)^2)=x^3f(xf(y)^2)=x^3y^3f(xy)$ (check the main equation)

so $f(f(x)^2f(y)^2)=f(f(xy)^2)\Longrightarrow f(x)^2f(y)^2=f(xy)^2$ (multiplicative function)

therefore :

$f(x)=0$

$f(x)=1$ (which is not possible)

$f(x)=x^a$ if we sub this in the main equation $\Longrightarrow 2a^2 = a+3 \Longrightarrow a=\frac{3}{2}$ or $a=-1$

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    The problems specifies $f:\mathbb{Q^+}\rightarrow \mathbb{Q^+}$, so only -1 is possible. – Ricky Aug 28 '23 at 17:12