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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 $$f\big(yf(x)\big)=x^2y^4$$

Mohsen Shahriari
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Vlad9pa
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2 Answers2

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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 $f(yf(x))=x^2y^4$

pointer
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  • why is The first sentence true? –  Nov 22 '14 at 09:49
  • Which part do you mean? – pointer Nov 22 '14 at 09:51
  • The first sentence of your answer. –  Nov 22 '14 at 09:52
  • There are 3 formulas in the first sentence. Which of them don't you agree with? – pointer Nov 22 '14 at 09:56
  • The third one I dont understant. –  Nov 22 '14 at 09:58
  • @PedroArbizu Putting $x=1$ in $f(yf(x))=x^2y^4$, we get $f(yf(1))=y^4$. If $f(1)=0$ this would mean $f(0)=y^4$ for all $y$, so $y^4$ is a constant, which is absurd; so $f(1)\ne0$. Replace $y$ with $y/f(1)$ in $f(yf(1))=y^4$. – bof Nov 22 '14 at 10:01
  • I understand now. I had to use some paper to see. Maybe, just to make it more understanble for the OP, you should do $z = y f(1) $. and then change $z$ for $y$. (+1). Apologies for being slow. –  Nov 22 '14 at 10:01
  • Putting $y=1$ in $f(y)=(y/f(1))^4$ we get $f(1)^5=1$, so $f(1)=1$ since $f$ is assumed to be real-valued. Could there be complex-valued solutions? – bof Nov 22 '14 at 10:04
  • This is a bad answer to the question asked, which was for a start. – Paul Nov 22 '14 at 10:14
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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, $f(yf(x)) = y^\sqrt{2} x^2$

David
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