Rational FEs

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#1 h Jul 17, 2019, 12:12 PM • 17 Y

Y by MarkBcc168, jisunxingfu, A-Thought-Of-God, centslordm, megarnie, jhu08, HWenslawski, ImSh95, Adventure10, Mango247, Spiritpalm, deplasmanyollari, Akacool, Sedro, Korean_fish_Kaohsiung, Funcshun840, ItsBesi

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 $x,y\in\mathbb{Q}_{>0}$

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Kayak

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Kayak

#2 h Jul 17, 2019, 12:25 PM • 6 Y

Y by centslordm, jhu08, megarnie, HWenslawski, ImSh95, Adventure10

Also Indian TST D2 P1

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v_Enhance

6870 posts

v_Enhance

#3 h Jul 17, 2019, 12:55 PM • 26 Y

Y by omerediprs, bubu_2001, Nathanisme, v4913, son7, Siddharth03, mijail, centslordm, math31415926535, megarnie, jhu08, HamstPan38825, myh2910, ImSh95, sail0r, fastandfuriuos, michaelwenquan, two_steps, Adventure10, sabkx, D_S, Sedro, Funcshun840, NicoN9, MS_asdfgzxcvb, NonoPL

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.

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TheDarkPrince

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TheDarkPrince

#4 h Jul 17, 2019, 1:04 PM • 18 Y

Y by Pluto1708, Arhaan, ayan_mathematics_king, ftheftics, Math-wiz, DeZade2002, bubu_2001, hsiangshen, microsoft_office_word, Pluto04, A-Thought-Of-God, centslordm, jhu08, HoRI_DA_GRe8, ImSh95, rayfish, Adventure10, NonoPL

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. $~\square$

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BlazingMuddy

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BlazingMuddy

#5 h Jul 17, 2019, 1:05 PM • 13 Y

Y by Dividend, pavel kozlov, Nathanisme, Pluto04, Siddharth03, centslordm, jhu08, megarnie, HoRI_DA_GRe8, ImSh95, chystudent1-_-, Adventure10, ihatemath123

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.

This post has been edited 2 times. Last edited by BlazingMuddy, Jul 17, 2019, 1:08 PM
Reason: more edits :(

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Leartia

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Leartia

#6 h Jul 17, 2019, 1:23 PM • 4 Y

Y by centslordm, HoRI_DA_GRe8, ImSh95, Adventure10

$f(x)=1, \forall x\in \mathbb{Q^+}$

$P(x,y) \rightarrow f(x^{2}f(y)^2)=f(x)^{2}f(y)$

$P(1,x) \rightarrow f(f(x)^2)=f(1)^{2}f(x)...(A)$

$P(\displaystyle \frac{x}{f(x^2)},x^2) \rightarrow f(x^2)=f(\frac{x}{f(x^2)})^{2}f(x^2)$ $\Rightarrow f(\displaystyle \frac{x}{f(x^2)})=1$

$\exists c\in \mathbb{Q^+}$ 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 $f(a)\neq 1$ .

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MarkBcc168

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MarkBcc168

#7 h Jul 17, 2019, 1:56 PM • 7 Y

Y by Aryan-23, Pluto1708, centslordm, ImSh95, GeoMetrix, Adventure10, Mango247

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 $x$ .

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sbealing

307 posts

sbealing

#8 h Jul 17, 2019, 5:56 PM • 4 Y

Y by centslordm, ImSh95, Adventure10, Mango247

$P \left(f(x),y \right) \Rightarrow f \left(f(x)^2 f(y)^2 \right)=f(f(x))^2 f(y)$ 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

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yayups

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yayups

#9 h Jul 18, 2019, 6:46 AM • 4 Y

Y by Pluto04, centslordm, ImSh95, Adventure10

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.

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lminsl

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lminsl

#10 h Jul 18, 2019, 8:34 AM • 6 Y

Y by kjy1102, Pluto04, centslordm, ImSh95, Adventure10, Mango247

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 $f \equiv 1$ .

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ubermensch

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ubermensch

#11 h Jul 18, 2019, 4:15 PM • 5 Y

Y by centslordm, ImSh95, Adventure10, Mango247, ATGY

$P(\frac{1}{f(y)},y): f(1)=f(\frac{1}{f(y)})^2f(y)$ 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 $f \equiv 1$

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niyu

830 posts

niyu

#12 h Jul 19, 2019, 3:00 AM • 3 Y

Y by centslordm, ImSh95, Adventure10

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. $\Box$

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mastermind.hk16

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mastermind.hk16

#13 h Aug 14, 2019, 7:34 PM • 3 Y

Y by centslordm, ImSh95, Adventure10

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.

This post has been edited 7 times. Last edited by mastermind.hk16, Aug 15, 2019, 3:11 AM

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rocketscience

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rocketscience

#14 h Aug 17, 2019, 10:47 PM • 4 Y

Y by Imayormaynotknowcalculus, centslordm, ImSh95, Adventure10

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.

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franchester

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franchester

#15 h Dec 23, 2019, 11:04 PM • 3 Y

Y by centslordm, ImSh95, Adventure10

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 $f(x)=1$ .

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GrantStar

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GrantStar

#108 h May 3, 2024, 1:41 AM

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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.

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sami1618

881 posts

sami1618

#109 h May 5, 2024, 8:58 PM

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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 $f(y)=1$ .

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ryanbear

1055 posts

ryanbear

#110 h Jun 22, 2024, 4:32 PM

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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 $y$ .

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joshualiu315

2513 posts

joshualiu315

#111 h Aug 31, 2024, 5:15 AM

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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.

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ezpotd

1251 posts

ezpotd

#112 h Sep 23, 2024, 2:24 PM

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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 $x$ .

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N3bula

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N3bula

#113 h Oct 11, 2024, 11:38 PM

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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 $y$ .

This post has been edited 1 time. Last edited by N3bula, Oct 12, 2024, 12:10 AM

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SimplisticFormulas

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SimplisticFormulas

#115 h Dec 8, 2024, 9:30 AM

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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, $f(y)=1$ $\forall y \in \mathbb{Q}_{>0}$ . $\blacksquare$
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L13832

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L13832

#116 h Dec 22, 2024, 5:09 AM

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Nice problem!
solution

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Sedro

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Sedro

#117 h Dec 27, 2024, 6:19 AM

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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 $y\in \mathbb{Q}_{>0}$ . $\blacksquare$

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SimplisticFormulas

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SimplisticFormulas

#118 h Dec 28, 2024, 8:51 AM

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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). $\blacksquare$

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Maximilian113

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Maximilian113

#119 h Jan 18, 2025, 4:45 AM

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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.

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Ilikeminecraft

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Ilikeminecraft

#120 h Jan 30, 2025, 6:04 AM

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cool!

$\left(\frac1{f(1)}, 1\right) \implies f(1) = f(\frac1{f(1)})^2 f(1).$ Since $f(x) > 0,$ we can conclude that $f\left(\frac1{f(1)}\right) = 1.$

$\left(x, \frac1{f(1)}\right) \implies f(x^2) = f(x)^2.$ Taking $x = 1$ into this equation implies $f(1) = 1.$
Claim: $\sqrt[2^k]{f(x)} \in \mathbb Q_{> 0}.$
Proof: Trivial by induction. Just apply the fact that the range for the previous $k$ value is all rationals to finish.
Clearly, this implies $f\equiv1.$

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ihatemath123

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ihatemath123

#121 h Mar 17, 2025, 4:07 AM

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Claim: We have $f(x) = 1$ for some $x$ .
Proof: We can make the first and third terms cancel with $P(\tfrac{t}{f(t)^2}, t^2)$ .

Taking $P(x,r)$ for some $r$ such that $f(r)=1$ gives us $f(x^2) = f(x)^2$ for all $x$ . So, we have
$\left( \frac{f(xf(y))}{f(x)} \right)^2 = f(y).$ This implies that $f(x)$ is a perfect square for all $x$ , which in turn implies that it is a perfect fourth power, and so on. This forces $f(x) = 1$ for all $x$ . :love:

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MuradSafarli

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MuradSafarli

#122 h Mar 28, 2025, 12:22 AM • 1 Y

Y by zaidova

My solution:

From the given functional equation, substituting $P(x / f(x^2), x^2)$ leads to $f(x / f(x^2)) = 1$ . This implies the existence of some $k$ such that $f(k) = 1$ .

Now, substituting $P(x, k)$ , we obtain:
$f(x)^2 = f(x^2) \quad (1)$ Using (1) in the original equation, we get:
$f(x f(y)) = f(x) f(\sqrt{y})$ Setting $x = f(x)$ and using symmetry, we derive:
$f(f(x)) f(\sqrt{y}) = f(f(y)) f(\sqrt{x})$ From this, we conclude that:
$f(f(x^2)) = c f(x) \quad (2)$ Now, applying $f()$ to both sides of the original equation, we obtain:
$f(f((x f(y))^2)) = f(f(x)^2 f(y))$ Using (2), this simplifies to:
$c f(x f(y)) = c f(f(x) \sqrt{f(y)})$ Setting $y = k$ , we get $f(x) = f(f(x))$ .

Finally, substituting this into (2), we obtain:
$f(f(x^2)) = c f(x) = f(x^2) = f(x)^2$ From this, it follows that $f(x) = c$ , and the only function satisfying the equation is $f(x) = 1$ .

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jl_

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jl_

#123 h Mar 29, 2025, 10:22 AM

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Notice that plugging $x=\frac{1}{f(1)}, y=1$ we get $f(1)=f(\frac{1}{f(1)})^2f(1),$ which tells us that $f(\frac{1}{f(1)})=1$ . Plugging $y=\frac{1}{f(1)}$ gives us $f(x^2)=f(x)^2,$ which gives us $f(1)=1$ . Then plug $x=1$ into the original equation to get $f(f(y)^2)=f(y).$ Now notice that if I plug in $y:=f(y)^2$ , I get $f(x^2f(y)^2)=f(x)^2f(y)^2=f(x)^2f(y),$ but that means $f(y)=1$ for all $y$ . $\blacksquare$

This post has been edited 3 times. Last edited by jl_, Mar 29, 2025, 10:25 AM

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