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It's my last question. Just give me advice how to start.

Find all such functions $$f:\mathbb R\to \mathbb R\text,$$ for all real $x$ and $y$, the equality $$f\big(yf(x)\big)=x^2y^4$$

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  • $\begingroup$ are the function continuous ? derivable ? $\endgroup$
    – idm
    Nov 22, 2014 at 9:27
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    $\begingroup$ Show that f(1) is not 0. $\endgroup$
    – Paul
    Nov 22, 2014 at 9:36
  • $\begingroup$ Function are contimuos. $\endgroup$
    – Vlad9pa
    Nov 22, 2014 at 9:36

2 Answers 2

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$ f(yf(1))=y^4, $ so $f(1)\ne0$, so $f(y)=\left(\frac{y}{f(1)}\right)^4$. Now putting $y=1$ in last equation we get $f(1)=1$ and $f(y)=y^4$. Now it is easy too see that $f(y)=y^4$ cannot satisfy the equation $f(yf(x))=x^2y^4$

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  • $\begingroup$ why is The first sentence true? $\endgroup$
    – user139708
    Nov 22, 2014 at 9:49
  • $\begingroup$ Which part do you mean? $\endgroup$
    – pointer
    Nov 22, 2014 at 9:51
  • $\begingroup$ The first sentence of your answer. $\endgroup$
    – user139708
    Nov 22, 2014 at 9:52
  • $\begingroup$ There are 3 formulas in the first sentence. Which of them don't you agree with? $\endgroup$
    – pointer
    Nov 22, 2014 at 9:56
  • $\begingroup$ The third one I dont understant. $\endgroup$
    – user139708
    Nov 22, 2014 at 9:58
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A bit late to the party, but:

In the case $y=1$, we get $f(f(x)) = x^2$, so $f(x)$, it it exists at all, must equal $x^\sqrt{2}$ for all possible $x$

But now we run into trouble as $f(x):=x^{\sqrt{2}}$ does not satisfy that property for y $\neq 1$. Indeed, $f(yf(x)) = y^\sqrt{2} x^2$

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