Easy for P3. Just prove that $f(f(f(x)))=x$ and plug in $y = f(f(y))$ and we're done.

here is my idea

pikapool wrote: Easy for P3. Just prove that $f(f(f(x)))=x$ and plug in $y = f(f(y))$ and we're done. Isn’t $f(f(f(x)))=f(x)$?

Solilin wrote: Let $c$ be a positive real number. Find all functions $f:\mathbb{R}^+\to\mathbb{R}^+$ that satisfy $$x^2f(xf(y))f(x)f(y)=c$$for all positive reals $x$ and $y$. $P(1,y)$ gives $f(f(y))f(y)=\frac{c}{f(1)}$ (1) $P(f(x),y)$ using (1) gives: $f(x)^2f(f(x)f(y))f(f(x))f(y)=c\Rightarrow f(f(x)f(y))f(x)f(y)=f(1)$ (2) $P(x,1)$ gives: $f(xf(1))f(x)=\frac{c}{f(1)x^2}$ so $f(a)f(b)$ is surjective. So in (2) if we set $z=f(x)f(y)$ we get :$f(z)=\frac{f(1)}{z}$ And now we get esily that $f(x)=\frac{\sqrt{c}}{x}$ is the only sollution

Plugging in $x=\tfrac y{f(y)}$ gives $f\left(\tfrac y{f(y)}\right) = \tfrac c{y^2}$, so $f$ is surjective. Now, take $y$ such that $f(y)=1$ to get $(xf(x))^2 = c$, or $f(x)=\tfrac{\sqrt c}x$, which satisfies the equation.

The only solution is $\boxed{f(x)=\frac{\sqrt{c}}{x}}$ which can easily be verified to work. Let the given assertion be $P(x,y)$. Claim: $f$ is surjective Consider the assertion $P\left(\frac{y}{f(y)},y\right)$ which yields $f\left(\frac{y}{f(y)}\right)=\frac{c}{y^2}$ finishing the claim. Claim: $f(x):\frac{1}{x}$ Consider the assertion $P(1,y)$ which yields $f(f(y))=\frac{c}{f(1)f(y)}$. Since $f$ is surjective the claim follows. Now we can simply check all functions of the form $f(x)=\frac{a}{x}$ for some positive real number $a$.

Let $P(x,y)$ denote the statement $x^2f(xf(y))f(x)f(y)=c$ Claim 1: $f$ is surjective Proof: $P\left(\frac{y}{f(y)},y \right): f\left(\frac{y}{f(y)} \right)=\frac{c}{y^2}$ Claim 2: $f$ is injective Proof: Suppose not. Then there exists $a$, $b>0$ such that $a \neq b$ and $f(a)=f(b)$ \[P(a,x): a^2f(af(x))f(a)f(x)=c\]\[P(b,x): b^2f(bf(x))f(b)f(x)=c\]Taking $x$ such that $f(x)=1$ by Claim 1, we get that $a^2=b^2$, and since we are working over positive reals, $a=b$. Claim 3: $c=f(1)^2$ Proof: $P(1,1): f(f(1))f(1)^2=c$ $P(1,f(1)): f(f(f(1)))f(f(1))f(1)=c$ Thus, $f(f(1))=1$, so $c=f(1)^2$. $P\left(\frac{1}{f(y)},y\right)=P\left(\frac{1}{f(y)},1\right): f(y)=f\left(\frac{f(1)}{f(y}\right) \implies f(y)=\frac{f(1)}{y}$, which works.