Also Indian TST D2 P1

The only answer is $f \equiv 1$. Claim: We have $f(x^2) = f(x)^2$ for all $x \in {\mathbb Q}_{>0}$. Proof. Letting $P(x,y)$ denote the given statement, plug in $P(f(1)^{-1}, 1)$ to deduce $f(f(1)^{-1}) = 1$. Then $P(x, f(1)^{-1})$ finishes. $\blacksquare$ Now, write the given as \[ f(y) = \left[ \frac{f(xf(y))}{f(x)} \right]^2. \]We prove by induction on $n \ge 0$ than all elements of $\operatorname{Img} f$ are $2^n$'th powers for each $n$. The base case $n = 0$ is vacuous, and the inductive step follows by above identity. It follows we must have $f$ equal to $1$ always. Okay, the end. Remark: Note that $x \mapsto \sqrt x$ works if ${\mathbb Q}_{>0}$ is replaced by ${\mathbb R}_{>0}$. So the codomain actually matters here.

Functional wrote: Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f:\mathbb{Q}_{>0}\to \mathbb{Q}_{>0}$ satisfying $$f(x^2f(y)^2)=f(x)^2f(y)$$for all $x,y\in\mathbb{Q}_{>0}$ Solution Plugging $x = \frac{1}{f(1)}, y = 1$ gives us $f\left(\frac{1}{f(1)}\right) = 1$. Plugging $x, \frac{1}{f(1)}$ gives us $f(x^2) = f(x)^2$. This gives us $f(xf(y))^2 = f(x)^2f(y)$ or $f(y)$ is a square of a rational for all positive rationals $q$. Now plug $f_1(x) = \sqrt{f(x)}$ to get $f_1(xf_1(y)^2)^4 = f_1(x)^4f_1(y)^2$ which again gives $f_1(y)$ is a square of a rational for all positive rationals $q$. This eventually gives us that $f(x)^{\frac{1}{2^n}}$ is a rational which is false for large $n$, therefore $f(x) = 1$ is the only solution which works. $~\square$

As said above, the only answer is $f\equiv 1$. By the original equality, for all $x, y\in \mathbb{Q}_{> 0}$, $$f(f(x))^2 f(y) = f(f(x)^2 f(y)^2) = f(f(y))^2 f(x)$$ i.e. $\frac{f(x)}{f(y)} = \left(\frac{f(f(x))}{f(f(y))}\right)^2$ for all positive rational numbers $x$ and $y$. Continuing by induction, $$ \frac{f(x)}{f(y)} = \left(\frac{f^k(x)}{f^k(y)}\right)^{2^{k-1}} $$ for all $k\geq 1$, where $f^k$ here means $k$-fold application of $f$. This implies that $2^{k-1}|v_p\left(\frac{f(x)}{f(y)}\right)$ for every positive integer $k$ and prime numbers $p$, so $f$ must be constant, and thus $f\equiv 1$, as desired.

$f(x)=1, \forall x\in \mathbb{Q^+}$ $P(x,y) \rightarrow f(x^{2}f(y)^2)=f(x)^{2}f(y) $ $P(1,x) \rightarrow f(f(x)^2)=f(1)^{2}f(x)...(A)$ $P(\displaystyle \frac{x}{f(x^2)},x^2) \rightarrow f(x^2)=f(\frac{x}{f(x^2)})^{2}f(x^2)$ $\Rightarrow f(\displaystyle \frac{x}{f(x^2)})=1$ $\exists c\in \mathbb{Q^+}$ such that $f(c)=1$ $P(x,c) \rightarrow f(x^2)=f(x)^2$ $P(x,y^2) \rightarrow f(xf(y)^2)^2=f(x)^{2}f(y)^{2}$ $\Rightarrow f(xf(y)^2)=f(x)f(y) ... (1)$ In (1) let $x=1$ and we obtain $f(f(y)^2)=f(1)f(y)...(B)$ Combining (A) and (B) yields $f(1)=1$ So $f(f(x)^2)= f(f(x))^2= f(x)$ $\Rightarrow f^k(x)=\sqrt[k]{f(x)}$ where $k$ is a power of $2$. Suppose that for some $a\in \mathbb{Q^+}$ such that $f(a)\neq 1$ let $f(a)=\displaystyle \frac{m}{n}$ such that $\gcd(m,n)=1$ Let $k$ be a power of $2$ with $V_{p}(m)<k \wedge V_{p}(n)<k$ for all primes $p$. $f^k(a)=\sqrt[k]{f(a)}$ so $\sqrt[k]{f(a)} \in \mathbb{Q^+}$ Let $\sqrt[k]{\displaystyle\frac{m}{n}}=\displaystyle\frac{x}{y}$ for some $x,y \in \mathbb{N}$ with $\gcd(x,y)=1$ $my^k=nx^k$ if $p \mid m \Rightarrow p \mid x$ $V_{p}(my^k)=V_{p}(nx^k)$ Since $\gcd(x,y)=1$ $V_{p}(y)=0$ $V_{p}(my^k)=V_{p}(m)=V_{p}(x^k)=k \cdot V_{p}(x) \geq k$ this is clearly a contradiction and thus there doesn't exist any positive rational number $a$ with $f(a)\neq 1$.

Only constant function $x\to 1$ works. It clearly does. As usual, let $P(x,y)$ denote the assertion $f(x^2f(y)^2) = f(x)^2f(y)$. $$P\left(\frac{1}{f(1)},1\right)\implies f(\text{something})=1.$$If $f(a)=1$, then $P(x,a)\implies f(x^2)=f(x)^2$. Thus $$P(1,x)\implies f(x) = \left(\frac{f(f(x))}{f(1)}\right)^2\qquad (*).$$If we denote $\mathbb{Q}^n$ by $\{q^n : q\in\mathbb{Q}\}$ then (*) implies $\mathrm{Im}(f) \subseteq \mathbb{Q}^2$. Using (*) repeatedly gives $\mathrm{Im}(f) \subseteq \mathbb{Q}^4$, $\mathrm{Im}(f) \subseteq \mathbb{Q}^8$, $\mathrm{Im}(f) \subseteq \mathbb{Q}^{16}$,... As $\bigcap\limits_{n=1}^\infty \mathbb{Q}^{2^n} = \{1\}$, this forces $f(x)=1$ for all $x$.

$$P \left(f(x),y \right) \Rightarrow f \left(f(x)^2 f(y)^2 \right)=f(f(x))^2 f(y)$$Swapping $x,y$ and equating gives: $$f(f(x))^2 f(y)=f(f(y))^2 f(x) \Rightarrow \frac{f(f(x))^2}{f(x)}=\frac{f(f(y))^2}{f(y)}=c$$for some constant $c \in \mathbb{Q}_{>0}$ (as this holds $\forall x,y$). We now claim: $$\underbrace{f \circ \cdots \circ f}_{\text{n times}}(x)=f^n(x)=c \cdot \left(\frac{f(x)}{c} \right)^{1/2^{n-1}}$$for all $n \geq 1$. This clearly holds for $n=1$. We go by induction. For $n \geq 2$: $$f^{n}(x)=f(f(f^{n-2}(x)))=\left(c f(f^{n-2}(x)) \right)^{1/2}=\left(c f^{n-1}(x) \right)^{1/2}=\left(c^2 \left(\frac{f(x)}{c} \right)^{1/2^{n-2}} \right)^{1/2}$$and expanding out shows the result is true by induction. But then we have: $$\left(\frac{f(x)}{c} \right)^{1/2^{n-1}}=\frac{f^{n}(x)}{c} \in \mathbb{Q}_{>0}$$for all $n$. But if $p$ is prime this means for all $n$: $$2^{n-1} \vert v_p \left(\frac{f(x)}{c} \right) \Rightarrow v_p \left(\frac{f(x)}{c} \right)=0 \Rightarrow \frac{f(x)}{c}=1$$Hence $f$ is constant and $c$ satisfies $c=c^3$ so we must have $c=1$ hence: $$\boxed{f(x) \equiv 1}$$which indeed works

ISL 2018 A1 wrote: Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f:\mathbb{Q}_{>0}\to \mathbb{Q}_{>0}$ satisfying $$f(x^2f(y)^2)=f(x)^2f(y)$$for all $x,y\in\mathbb{Q}_{>0}$ Let $P(x,y)$ denote the given FE. Let $k=f(1)\in\mathbb{Q}_{>0}$. Now, \[P(x,1)\implies f((xk)^2)=f(x)^2k.\]Plugging in $x=1/k$ into this, we get that $k=f(1/k)^2k$, so $f(1/k)=1$. Now, \[P(x,1/k)\implies f(x^2)=f(x)^2.\]Thus, the FE implies \[f(y)=\left(\frac{f(xf(y))}{f(x)}\right)^2.\] Claim: $f(y)$ is always a perfect $2^k$th power. Proof of Claim: We proceed by induction on $k$. The case $k=0$ is trivial. Now, the equation we derived above shows that \[f(y)=a^{2^{k+1}}\]for some rational $a$, which completes the induction. $\blacksquare$ This lemma then trivially implies that $f(x)=1$ for all $x$, which can easily be checked to be a solution.

Note that the only element $q\in \mathbb{Q}_{>0}$ which is a rational number when taken $\frac{1}{2^k}$-th power is $1$. Let $x=f(z)$ at the given assertion, then $f(f(z)^2f(y)^2)=f(f(z))^2f(y)=f(f(y)^2)f(z)$. Hence there exists some $c \in \mathbb{Q}_{>0}$ such that $\frac{f(f(x))^2}{f(x)}=c$ for any $x$. Now consider $g(x)=\frac{f(x)}{c}$, then $g(f(x))^2=g(x)$ for any $x$. This means that $g(x), \sqrt{g(x)}, \cdots, \sqrt[2^k]{g(x)}, \cdots$ are all rationals, giving that $g$ is constantly $1$. Hence $f$ is constant, so $f \equiv 1$.

$P(\frac{1}{f(y)},y): f(1)=f(\frac{1}{f(y)})^2f(y)$ and $P(\frac{1}{f(y)^2},y): 1=f(\frac{1}{f(y)^2})f(y)$. Put $y=1$ to get $f(1)=f(\frac{1}{f(1)})^2f(1)=> f(\frac{1}{f(1)})=1$. Now $P(x,\frac{1}{f(1)}): f(x^2)=f(x)^2$. $P(1,y): f(f(y)^2)=f(y)=>f(f(y))=\sqrt{f(y)}$ Now, $P(\frac{1}{f(y)},f(y)): f(\frac{1}{f(y)^2} \cdot f(y))=f(\frac{1}{f(y)})^2 \cdot \sqrt{f(y)} => f(\frac{1}{f(y)}) \cdot \sqrt{f(y)} =1=>$$f(\frac{1}{f(y)})=\frac{1}{\sqrt{f(y)})}$. Let's take $k=f(y)$, where $k$ is any value which $f(y)$ can take. We know $f(kx)=\sqrt{k} f(x)$. Putting $x=\frac{1}{\sqrt{k}}$, we get $f(\sqrt{k})=\sqrt{k} \cdot f(\frac{1}{f(k)})$(as $f(k)=\sqrt{k}$)$=>f(k^{\frac12})=k^\frac14$(thus $k^{\frac14}$ is rational). Now repeat the drill to get $k^{\frac{1}{2^n}}$ is rational $=> k=1$. $=>$ Only solution is $f \equiv 1$

We claim the answer is $\boxed{f \equiv 1}$, which obviously works. Let $c = f\left(1\right)$, and let $P\left(x, y\right)$ denote the initial assertion. From $P\left(1, 1\right)$, we have \begin{align*} f\left(c^2\right) &= c^3. \end{align*}Now, from $P\left(c^2, 1\right)$, we have \begin{align*} f\left(c^6\right) &= c^7, \end{align*}while from $P\left(1, c^2\right)$, we have \begin{align*} f\left(c^6\right) &= c^5. \end{align*}Hence, $c^5 = c^7$, and since $c \in \mathbb{Q}_{> 0}$, $c = 1$. Therefore, $f\left(1\right) = 1$. Now, $P\left(x, 1\right)$ implies \begin{align*} f\left(x^2\right) &= f\left(x\right)^2. \end{align*}From $P\left(1, x\right)$, we have \begin{align*} f\left(f\left(x\right)^2\right) &= f\left(x\right) \\ f\left(f\left(x\right)\right)^2 &= f\left(x\right). \end{align*}Therefore, if $k$ is in the image of $f$, so is $k^{\frac{1}{2}}$. Iterating, we find that in fact $k^{\frac{1}{2^\ell}}$ is in the image of $f$ for all positive integers $\ell$. Now, suppose $k \neq 1$. Then, for $\ell$ large enough, $k^{\frac{1}{2^\ell}}$ will not be rational, contradiction. Hence, we must have $k = 1$. This immediately implies that $f \equiv 1$ identically, concluding the proof. $\Box$

The answer is $f(x) =1 \ \ \forall x \in \mathbb{Q^+}$. Plug $x=\frac{1}{f(1)^2}\overset{def}{=} t, y=1 \longrightarrow f(1)=f(t)^2 f(1) \longrightarrow f(t)=1$. Then $y=t \longrightarrow f(x^2) = f^2(x) \ \ \forall x$. From here, $f(1)= f^2(1) \longrightarrow f(1)=1$. Plug $x=1 \longrightarrow f^2(f(y))=f(f^2(y))=f(y) \longrightarrow f(f(y)) = \sqrt{f(y)} \dots (1)$. Fix a rational number $l$. Claim: $f(l)$ is a $2^{n}$ power of a rational for all $n \in \mathbb{N}$. Let $f^{[n]}(x) = \underbrace{f(f(\dots f(x))\dots )}_{n \ times}$ Then iterating $(1)$ we see that $$f^{[n+1]}(l) = f^{[n]}(f(l))= (f(f(l))^{\frac{1}{2^{n-1}}}=(f(l))^{\frac{1}{2^n}}$$And so our claim is proved. Now, suppose $f(l) \neq 1$. Then by our claim, at some point $f(l)$ is irrational. Contradiction. So we are done.

As usual, let $P(x, y)$ denote the assertion and let $R$ denote the range of $f$. Observe that $P(1/f(1), 1)$ gives $1 \in R$. Then taking $P(x, c)$ where $f(c) = 1$ gives $f(x^2) = f(x)^2$. As a corollary, $f(1)=1$. Now $P(x, y)$ rewrites as \[f(xf(y)) = f(x) \sqrt{f(y)},\]and letting $x=1$ gives $f(f(y)) = \sqrt{f(y)}$. Equivalently, $c \in R$ implies $\sqrt{c} \in R$. It follows that $R = \{1\}$, so the constant function is our only solution.

Let $P(x,y)$ denote the given assertion. I claim that $f(x)=1$ is the only solution, which is easily verifiable, and I will do this by showing that $f(x)$ is a perfect $2^n$th power of a rational for all $x\in \mathbb{Q}_{>0}$. We proceed with induction on $n$. The base case is easy: taking $P(1,x)$ gives $$f(f(x))^2=f(1)^2f(x)\implies f(x)=\left(\frac{f(f(x))}{f(1)}\right)^2$$Since $f$ maps $\mathbb{Q}_{>0}$ to $\mathbb{Q}_{>0}$, we must have $f(x)$ is a perfect square of a rational number for all $x\in \mathbb{Q}_{>0}$. Now assume the claim is true for some $n=k\geq 1$. Because of the assumption, we can set a function $g(x)$ such that $g(x)^{2^k}=f(x)$ over the same domain and range. We have $$g^{2^k}(x)=\left(\frac{g^{2^k}(f(x))}{g^{2^k}(1)}\right)^2=\left(\frac{g(f(x))}{g(1)}\right)^{2^{k+1}}$$Taking the $2^k$th root gives $$g(x)=\left(\frac{g(f(x))}{g(1)}\right)^2$$Since $g(x)$ is over $\mathbb{Q}_{>0}$, the LHS must be a perfect square of a rational for all $x$. Due to the definition of $g$ this means that $f(x)$ is a perfect $2^{k+1}$th power of some function over the positive rationals as well, completing the induction. Since $f(x)$ is a perfect $2^n$th power for any positive integer $n$, we must have that $f(x)=1$.

Let $f(1)=c$ and let $P(x,y)$ denote the problem statement. $$\text{ (1) } P(1,y): f(1^2f(y)^2)=f(1)^2f(y) \implies f(f(y)^2)=c^2f(y) $$$$ \text{ (2) } P(x,1): f(x^2f(1)^2)=f(x)^2f(1) \implies f(c^2x^2)=cf(x)^2$$ From $(2)$ $$f(1)=cf(\frac{1}{c})^2 \implies c=cf(\frac{1}{c})^2 \implies f(\frac{1}{c})=1$$ $$P(x, \frac{1}{c}): f(x^2f(\frac{1}{c})^2)=f(x)^2f(\frac{1}{c}) \implies f(x^2)=f(x)^2$$$$P(\frac{1}{c}, y): f((\frac{1}{c})^2f(y)^2)=f(\frac{1}{c})^2f(y) \implies f(\frac{f(y)}{c})^2=f(y)$$so for all $y \in \mathbb{Q}_{>0}$, $f(y)$ is a perfect square of some $a \in \mathbb{Q}_{>0}$. However, tif $f(y) \neq 1$, $f(\frac{f(y)}{c}) \neq 1$ also, and is also a perfect square. By repeatedly iterating this and we can keep iterating this process until we get $f(a) \notin \mathbb{Q}_{>0}$, a contradiction. Therefore, $f(x)=1$ is the only solution.

proposed by Arnaud Maret. the french guy

Functional wrote: Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f:\mathbb{Q}_{>0}\to \mathbb{Q}_{>0}$ satisfying $$f(x^2f(y)^2)=f(x)^2f(y)$$for all $x,y\in\mathbb{Q}_{>0}$ Let $P(x, y)$ be the assertion. $P\left ( \frac{1}{f(1)}, 1 \right ) \implies RHS = f\left (\frac{1}{f(1)} \right ) = \frac{f(1)}{f(1)}$, and so $f^{-1}(1)$ is a unique number. Hence, we use $P\left (x, f(1)^{-1}\right )$ to obtain that $f(x^2) = f(x)^2$. So, we get that $f(xf(y))^2 = f(x)^2f(y) \implies f(y) = t^2$, where $t$ is a rational, and we see that if $f_t (x) = f(x) ^{\frac{1}{2^t}}$, we get that $f^{-t}(y)$ is still a square of a rational, and this will fail for a large enough $t$ unless and until $f(x) = p >0$, $p^{\frac{1}{2^t}}= p$ which gives $\boxed{f(x)=1}$ as the only solution.

Functional wrote: Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f:\mathbb{Q}_{>0}\to \mathbb{Q}_{>0}$ satisfying $$f(x^2f(y)^2)=f(x)^2f(y)$$for all $x,y\in\mathbb{Q}_{>0}$

Claim: The only solution is $f(x)\equiv 1$. Let $P(x,y)$ denote the assertion: $$f(x^2f(y)^2)=f(x)^2f(y)$$$P\left( \frac{1}{f(1)},1\right) \implies f\left( \frac{1}{f(1)} \right)=1 \implies \exists \alpha$ such that $f(\alpha)=1$. $$P(x,\alpha) \implies \boxed{f(x^2)=f(x)^2} \cdots(i)$$ $P(1,x^2) \implies f(1f(x)^2)^2=f(1)^2f(x)^2\implies f(f(x)^2)=f(1)f(x)$ $P(1,x) \implies f(f(x)^2)=f(1)^2.f(x)$ From these two equations, we get that $f(1)=1$. $P(1,x) \implies f(x)=f(f(x))^2$ Let $F_i=f(f(f(..._{\text{i times}}))$ Therefore, $\forall x$, we have $f(x)=F_2(x)^2$. Substituting this value of $f(x)$ in the original FE, we get: $$F_2(x^2F_2(y)^2)=F_2(x)^2F_2(y)$$Doing same thing recursively, we get that $F_2(x)=F_4(x)^2 \implies f(x)=F_4(x)^{2^2}$ Continuing by infinite descent, we gradually get that $f(x)=F_{2^i}(x)^{2^i}$ for all $i \in \mathbb{N}$ Thus, we can have no other solution, other than $\boxed{f(x)\equiv 1}$, and we're done!

Let the assertion be $P(x,y)$. We claim the answer is $f(x)=1$, which clearly works. We now show it is the only possible solution. Claim: $f\left(\frac{1}{f(x)}\right)=\sqrt{\frac{f(1)}{f(x)}}$ Proof: Using $P\left(\frac{1}{f(y)},y\right)$, we get: \[f(1)=f\left(\frac{1}{f(y)}\right)^2\cdot f(y),\]as desired $\square$ Claim: $f(x^2)=f(x)^2$ Proof: We now use $P(1,y)$, which can be written as: \[f(f(y)^2)=f(1)^2\cdot f(y).\]By the previous claim, this is also: \[f(f(y)^2)=\sqrt{f(1)}\cdot f(y).\]Therefore: \[f(1)^{3/2}=1,\]\[f(1)=1.\]Then, if we use $P(x,1)$, we get the desired result $\square$ When we substitute this back, we get: \[f(xf(y))^2=f(x)^2\cdot f(y).\]This means that $f(y)=g(y)^2$, such that $g:\mathbb{Q}_{>0}\to \mathbb{Q}_{>0}$. Applying this, we get: \[g(xg(y)^2)^4=g(x)^4g(y)^2,\]\[g(xg(y)^2)^2=g(x)^2g(y).\]As a result, $g$ must also be a square of a function $q: \mathbb{Q}_{>0}\to \mathbb{Q}_{>0}$, leading to an infinite descent. Therefore, $f$ must be a perfect $2^i$ power as $i$ approaches infinity, implying that $f$ must be $1$ $\blacksquare$

Substituting $(x,y)=(1/f(1),1)$ gives us $f(1/f(1))=1$. Now $(x,y)=(1,1/f(1))$ gives us $f(1)=1$. Now it is easy to get $f(x)^2=f(x^2)$ and $f(f(x))=f(x)^{1/2}$. It follows that $f^{k+1}(x)=f(x)^{2^{-k}}$ for all $k\in \mathbb N$ (using the notation of function composition), thus for $f$ to be rational it must be $1$. (Easy to check $f(x)=1$ for all $x$ satisfies the equation).

Let $P(x,y)$ denote the assertion $f(x^2f(y)^2)=f(x)^2f(y)$. Call a rational number $x$ an $n$-th power if $x=y^n$ for some rational number $y$. $P(\frac{1}{f(1)},1) \implies f(1)=f(\frac{1}{f(1)})^2f(1) \implies f(\frac{1}{f(1)})=1$. $P(1,\frac{1}{f(1)}) \implies f(1)=f(1)^2 \implies f(1)=1$. $P(x,1) \implies f(x^2)=f(x)^2$. We claim that for all positive integers $n$, $f(x)$ is a $2^n$-th power for all $x$. We will prove this by induction. $P(\frac{1}{f(y)},y) \implies 1=f(\frac{1}{f(y)})^2f(y) \implies f(y)$ is a 2nd power for all $y$. Now assume $f(x)$ is a $2^k$-th power for all $x$. $P(x,y) \implies f(y)=\frac{f(xf(y))^2}{f(x)^2} \implies f(y)$ is a $2^{k+1}$-th power. This proves the claim. It follows that $f(x)=1$ for all $x \in \mathbb{Q}_{>0}$. It is easy to check that this works.

Let $P(x,y)$ be the assertion. Note that $P\left(\frac{r}{f(r)},r^2\right)$ gives $f( \text{something} ) = 1$. Then, taking $f(y)=1$ we get $f(x^2)=f(x)^2$. To finish, $x=f(y)$ gives $f(f(y))^2=f(y)$, so if $r$ is in the range of $f$, then $f(r)^2=r$ so $\sqrt r$ is also, meaning eventually $r=1$. thus $f \equiv 1$ is the only solution which clearly works.

The only solution is $\boxed{f(x)=1}$ which can be easily verified to work. Claim: $f(x^2)=f(x)^2$ Plug in $(x,y)=(f(1)^{-2},1)$ giving $f(f(1)^{-2})=1$. Plug in $(x,y)=(x,f(1)^{-2})$ and the result follows. Now rewrite the assertion as, $$\frac{f(xf(y))^2}{f(x)^2}=f(y)$$This implies that $f(y)$ is a second power. If $f(y)$ is a perfect $2^{{n^{th}}}$ power then $f(y)$ must be a perfect $2^{{n+1^{th}}}$ power. Thus the only solution is $f(y)=1$.

Let $P(x,y)$ be the process of plugging in numbers for $x$ and $y$, respectively Let $f(1)=c$ $P(\frac{1}{c},1)$ to get that $c=f(\frac{1}{c})^2c$ $f(\frac{1}{c})=1$ $P(y,\frac{1}{c})$ to get that $f(y^2)=f(y)^2$ $P(1,\frac{1}{c})$ to get that $c=c^2$, so $c=1$ $f(1)=1$ $P(1,y)$ to get $f(f(y)^2)=f(y)=f(f(y^2))$ $P(f(y),y)$ to get $f(f(y)^4)=f(f(y)^2)^2=f(f(y))^4=f(f(y))^2f(y)$ $f(y)=f(f(y))^2$ Let $f(a)=b$ $f(a)=f(f(a))^2=f(b)^2=b$ $f(b)=\sqrt{b}$ $f(\sqrt{b})=\sqrt[4]{b}$ ... If this goes on forever and $b \neq 1$, then one of the outputs of the function will eventually become irrational. This is a contradiction, so that means $\boxed{f(y)=1}$ for all $y$.

The answer is $f(x) = \boxed{1}$, which is trivially shown to satisfy the given condition. Denote the given assertion as $P(x,y)$. Set $x = \tfrac{1}{f(1)}$ and $y = 1$: \[f(1) = f\left( \frac{1}{f(1)^2}\right)f(1)\]\[\implies f\left( \frac{1}{f(1)^2}\right) = 1.\] Then, $P(x,\tfrac{1}{f(1)^2})$ gives \[f(x^2) = f(x)^2.\] Plug this into the original equation to see that \[f(xf(y))^2 = f(x)^2f(y)\] and then let $x = f(y)$, which simplifies to \[f(f(y))^2 = f(y).\] Hence, if $b$ is in the range of $f$ such that $f(a)=b$, we see that $f(f(a))^2 = f(b)^2 = f(a) = b$, so $\sqrt{b}$ is also in the range of $f$; infinite descent finishes.

I claim the only answer is $f(x) = 1$ for all $x$, it is trivial to verify this works. Claim: Some $x$ exists such that $f(x) = 1$. Proof: Set $y = k^2$, then $x = \frac{k}{f(k^2)}$, so $f(x) =1$. Claim: If there exists some $x$ with $f(x) = a$, there exists some $y$ with $f(y) = \sqrt{a}$. Proof: Set $y = 1$. Finish: If there exists some $a$ with $\nu_p(a) \neq 0$ for at least one prime $p$, then we would see $a^{\frac{1}{2^k}}$ is in the range of $f$ for all $k$, but for large enough $k$ we would have $\nu_p(a) \equiv 1 \mod 2$, meaning we would not have a rational for $a^{\frac{1}{2^{k + 1}}}$, so such an $a$ cannot exist. Thus $1$ is the only possible element of the range of $f$, so $f(x) = 1$ for all $x$.

Let $P(x, y)$ denote the assertion. $P(\frac{1}{f(1)}, 1)$ \[f(\frac{1}{f(1)})=1\]Let $\frac{1}{f(1)}=k$. $P(x, k)$ \[f(x^2)=f(x)^2\]Thus we get that $f(y)=\frac{f(xf(y))^2}{f(x)^2}$, this implies that for $y$ $f(y)$ is the square of a rational number, however by repeating this argument we in fact see that for all $y$ $f(y)$ is a rational number raised to any abitrariy power of $2$ implying $f(y)=1$ for all $y$.

We shall prove that $f(y)=1$ $\forall y \in \mathbb{Q}_{>0}$. Let $P(x,y)$ denote the assertion $P \left(\frac{y}{f(y^2)}, y^2 \right)$ gives us that $f \left( \frac{y}{f(y^2)} \right) =1$ $P \left(x, \frac{y}{f(y^2)} \right)$ gives us that $f(x^2)=f(x)^2$ $P\left(\frac{y}{f(y^2)} ,y \right) $ gives us that $f \left(\frac{y}{f(y)} \right)^2 =f(y)$ Now this implies that $f(y)$ is a perfect square, but the; so is $f \left(\frac{y}{f(y)} \right)$, which would imply that $f(y)$ is also a perfect fourth power, and so on. Therefore, $f(y)=1$ $\forall y \in \mathbb{Q}_{>0}$. $\blacksquare$

Nice problem!

We claim that the only solution is $f \equiv 1$, which obviously works. We now show that it is the only one; let $P(x,y)$ denote the assertion. By $P(\tfrac{1}{f(1)},1)$, we have $f(1)=f(\tfrac{1}{f(1)})^2f(1)$, or $f(\tfrac{1}{f(1)})=1$. Let $u = \tfrac{1}{f(1)}$; note $f(u)=1$. By $P(x,u)$, we have $f(x^2)=f(x)^2$. Hence, the original assertion can be re-written as $f(xf(y))^2 = f(x)^2f(y)$. Thus, for any valid $y$, $f(y)$ is the square of a rational number. Suppose $f(y)$ is the $2^n$-th power of a positive rational number for any valid $y$. Then, $f(xf(y))^2$ and $f(x)^2$ are $2^{n+1}$-th powers of a rational number, which implies that $f(y)$ must be as well. Thus, $f(y)$ is equal to an arbitrarily large power of a positive rational for any valid $y$, which forces us to have $f(y)=1$ for every $y\in \mathbb{Q}_{>0}$. $\blacksquare$

cool We claim that $f(x)=1$ is the only such function. Let $P(x,y)$ denote the assertion. $P(x,x) \implies f(x^2 f(x)^2)=f(x)^3$ $…(a)$ $P\left( \frac{y}{f(y^2)},y^2 \right) \implies f\left( \frac{y}{f(y^2)} \right)=1$ $P\left( x,\frac{y}{f(y^2)} \right) \implies f(x^2)=f(x)^2$ $…(b)$ Now $(a)$ and $(b)$ give us that $f(xf(x))^2=f(x)^3$. Consider a function $f_{1}(x)$ such that $f(xf(x))^2=f(x)^3=f_{1}(x)^6$. Then we get $f_{1}(xf_{1}( x^2))^2=f_{1}(x)^3$. Consider $f_{2}(x)$ with $f_{1}(xf_{1}( x^2))^2=f_{1}(x)^3=f_{2}(x)^6$. Again we get $f_{2}(xf_{2}(x^4))^2=f_{2}(x)^3$ and so on. Hence $f(x)=f_{1}(x)^2=f_{2}(x)^4=…$ and $f(x)=1$ is the only such function (which indeed does satisfy the conditions). $\blacksquare$

Solved with a friend. Let $P(x, y)$ denote the assertion, and let $k=f(1).$ Then \begin{align*} P(1, x) &\implies f(f(x)^2) = k^2f(x) \\ P(f(x), 1) &\implies f(f(x)^2) = f(f(x))^2k \\ &\implies k^2f(x)=f(f(x))^2k \\ x=k^2 &\implies f(k^3)=k^2. \\ \end{align*}However, note that $P(1, 1) \implies f(k^2)=k^3.$ Hence, $P(1/k^2, f(k^3)) \implies k=f(1/k^2)^2k^2,$ thus $k$ is the square of a rational number. Therefore, $$P(\sqrt{k}, 1) \implies f(\sqrt{k})=\sqrt k.$$Hence $P(k, \sqrt{k}) \implies k=1.$ So we have found that $f(1)=1.$ Thus $$P(x, 1) \implies f(x^2)=f(x)^2 \implies P(1, x) \implies f(f(x))^2=f(x),$$so $f(x)$ is the square of a rational, say $d.$ Then the equation reduces to $f(f(d))^2=d,$ and we can continue this infinitely. But $f(x)$ cannot be a perfect $k$th power for arbitrarily large $k,$ unless it is $1.$ Therefore $f(x) \equiv 1$ is the only solution.