Denote by $\mathbb{Q}^+$ the set of all positive rational numbers. Determine all functions $f : \mathbb{Q}^+ \mapsto \mathbb{Q}^+$ which satisfy the following equation for all $x, y \in \mathbb{Q}^+:$ \[f\left( f(x)^2y \right) = x^3 f(xy).\] Proposed by Thomas Huber, Switzerland
Problem
Source: IMO Shortlist 2010, Algebra 5
Tags: function, algebra, IMO Shortlist, functional equation
17.07.2011 18:24
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=412941
16.03.2014 10:20
Sketch of proof. Let $g(x) = \frac{f(x)^2}{x}$. Then we have $g(g(x)y)g(x)=x^6g(y)$. Then $g(g(1))=1$. Also $g(g(x))g(x)=x^6g(1)$ and $g(g(x)x)=x^6$. Easy to get that $g$ is bijective. Thus we get $g$ is multiplicative. Hence $f$ is multiplicative. Hence we are left to solve \[f(f(x))^2=x^3f(x)\] where $f$ is injective and multiplicative. Define $\tau(x) = xf(x)$. We get $\tau(x)^3=\tau(f(x))^2\ldots (\ast)$. $\tau$ can be seen as a homomorphism from the group of rationals under multiplication to itself. By $(\ast)$ we get for all $x$, $\tau(x)=r^2$ for some rational $r$. Thus the image of the injective homomorphism is the squares of rationals$\ldots (1)$. But we get $\tau(f(x))=s^3$ for some rational $s$. Whereas $\tau(f(x))=\tau(r^2)=\tau(r)^2$. But that means for $\tau$ maps each rational to a $6$-th power of the rational. Which contradicts statement $(1)$. And hence $\tau$ must be equal to $1$. Solution \[\boxed{f(x)=\frac{1}{x}}\]
05.06.2014 14:30
let $p(x,y)=f\left( f(x)^2y\right) = x^3 f(xy)$ $y=1\Rightarrow f(f(x)^{2})=x^{3}f(x)\Rightarrow f$ is injective. $p(1,1) \Rightarrow f(1)=1$ $p(x,f(y)^{2})\Rightarrow f((f(x)f(y))^{2})=x^{3}y^{3}f(xy) (1)$ $ f(f(x)^{2})=x^{3}f(x) (2)$ $(1),(2)\Rightarrow f((f(x)f(y))^{2})=x^{3}y^{3}f(xy)=f((f(xy))^{2})\Rightarrow f(xy)=f(x)f(y)$ so $f(xy)=f(x)f(y) (3)$ $p(x,y),(3)\Rightarrow f(f(x))^{2}=x^{3}f(x) (4)$ let $g:\mathbb{Q}^{+}\rightarrow \mathbb{Q}^{+}$ is a function such that $g(x)=xf(x)$ $(3)\Rightarrow g(xy)=g(x)g(y) (5)$ $(4),(5)\Rightarrow g(g(x))^{2}=g(x)^{5} (6)$ from $(6)$ we find $\forall x\in \mathbb{Q}^{+}:g(x)=a^{2},a\in \mathbb{Q}^{+}\Rightarrow g(x)=g_{1}(x)^{2},g_{1}:\mathbb{Q}^{+}\rightarrow \mathbb{Q}^{+}$ so if we do it infinity time we find that $g(x)=1\Rightarrow f(x)=\frac{1}{x}$
08.01.2015 17:45
iarnab_kundu wrote: Also $g(g(x))g(x)=x^6g(1)$ and $g(g(x)x)=x^6$. Easy to get that $g$ is bijective. Why ?
14.04.2015 02:44
By $y=1$ we easily see $f$ is injective and by $x=1$ we see $f(1)=1$. With $y=1/x$ and $y=1/f(x)^2$ we get $f\left( \frac{f(x)^2}{x} \right) = x^3$ and $f\left( \frac{x}{f(x)^2} \right) = 1/x^3$. But also by injectivity $f(x/f(x)^2)=1/x^3 = f(f(1/x)^2/(1/x)) \Rightarrow x/f(x)^2 = f(1/x)^2/(1/x) \Rightarrow 1/f(x)=f(1/x)$ (1) Let $\Omega = \{ f(x)^2/x \}$, then if $a/b \in \Omega \Rightarrow f(a)/f(b)=f(a/b)$ (this is the problem statement), and therefore by (1) $f(ab)=f(a)f(b)$ for all $ab \in \Omega$ and $f(wa)=f(w)f(a)$ if $w \in \Omega$ by (1). By $x=a/f(a)^2,y=a^5$ in the original equation we get, since $x \in \Omega$, $1/f(a) = f(1/a)=f(f(x)^2y)=x^3f(xy)=(a^3/f(a)^6)f(x)f(y) = f(y)/f(a)^6 \Rightarrow f(a^5)=f(a)^5$ for all $a$. Now if $y=1$ then $f(f(x)^2)=x^3f(x)$ and so $\Omega \ni f(f(x)^2)^2/(f(x)^2) = x^6$ and thus any sixth power is in $\Omega$. Thus $f(a^6)=f(a^5)f(a)=f(a)^6$. So $f(a^3)f(a^3)=f(a^6)=f(a)^6$ and so $f(a^3)=f(a)^3 = f \left( \frac{f(f(a))^2}{f(a)} \right)$ and so $ a^3 \in \Omega$. Hence $f(a)^3=f(a^3)=f(a)f(a^2) \Rightarrow f(a^2)=f(a)^2$. Now by $y=1/f(x)$, $f(f(x))=x^3f(x/f(x))$ and by $y=1$, $x^3f(x)=f(f(x)^2)=f(f(x))^2=(x^3f(x/f(x)))^2$ so $f(x)=x^3f(x/f(x))^2=x^3f(x^2/f(x)^2)=x^3f((x^2/f(x))*(1/f(x)))=x^3f(x^2/f(x))f(1/f(x))=f(1/f(x))=1/f(f(x))$ and so $f(x)f(f(x))=1$ for all $x$. So if $\Gamma=\{ f(x) \}$ then $f(t)=1/t$ for all $t \in \Gamma$. Finally notice $x^3f(x)=f(f(x)^2)=f(f(x^2))=1/f(x^2)=1/f(x)^2$ and so $x^3f(x)^3=1$ and so $\boxed{f(x)=1/x}$.
10.07.2016 13:07
$f$ is injective : $f(a) = f(b) \implies f(f(a)^2) = f(f(b)^2) \implies a^3 f(a) = b^3 f(b) \implies a=b $. Now the problem is done by the following two lemmas: Lemma 1: f is multiplicative. proof: Putting $x=a$ and $y=f(b)$, we get $$ f(f(a)^2f(b)^2) = a^3f(f(b)^2 a) = (ab)^3 f(ab) $$, on the other hand substituting $x=ab$ and $y=1$, we deduce; $$ f(f(ab)^2) = (ab)^3 f(ab) $$Now these and injectivity yields the claim. Lemma 2: If $p \in \mathbb{Z}^{+}$ is a prime number, then $f(p) = \frac{1}{p}$ . proof: we know from fundamental theorem of arithmetic that any positive rational number can be written as product of integer powers of primes, we will say $x \in \mathbb{Q}^{+}$ to be free of prime $p$ if the power of such prime is zero in this decomposition. Let $f(p) = r $. Putting $x=r \ , \ y=1$, and using the above Lemma, we find $f(r)^2 = f(f(p))^2 = p^3 r $, then one can write $r = p^{2i+1} s^2 $, where $s \in \mathbb{Q}^{+}$ is free of $p$. So one gets $$ p^{(4i+2)(2i+1)} s^4 f(s)^4 = f(p^{2i+1} s^2)^2 = f(r)^2 =p^3 r = p^{2i+4} s^2 \implies f(s)^4 = p^{(4i+2)(2i+1)-(2i+4)} s^2$$. let $s = t^{2^k}$, where $t$ is not a rational square. Suppose $q$ is a prime appearing the factorization of $s$ ($q$ cannot equal $p$), then $2^{k+1}||\nu_q(s)$, but $\nu_q(f(s)^4) = \nu_q(f(t))^{2^{k+2}} = 2^{k+2} \nu_q(f(t)) $, so we arrive at a contradiction as the valuations of both sides don't agree. Thus $s=1$, hence $ 1= p^{(4i+2)(2i+1)-(2i+4)} $, which implies $i = -1$, that's $r = p^{(2i+1)}s^2 = p^{-1}$ as required. Combining the two lemmas , we get $f(x) = \frac{1}{x}$
29.11.2016 10:15
IMO ShortList 2010 A5 wrote: Denote by $\mathbb{Q}^+$ the set of all positive rational numbers. Determine all functions $f : \mathbb{Q}^+ \mapsto \mathbb{Q}^+$ which satisfy the following equation for all $x, y \in \mathbb{Q}^+:$ \[f\left( f(x)^2y \right) = x^3 f(xy).\]
04.10.2018 19:35
My solution: Let $P(x,y)$ denote the given assertion. Lemma 1 $f$ is injective. PROOF: Suppose $f(a)=f(b)$. Then $P(a,1)$ and $P(b,1)$ give $a^3=b^3$, i.e. $a=b$. Thus, $f$ is injective. $\Box$ Lemma 2 $f$ is multiplicative, i.e. $f(xy)=f(x)f(y) \text{ } \forall x,y \in \mathbb{Q^+}$. PROOF: $P(y,f(x)^2) \Rightarrow f(f(x)^2f(y)^2)=y^3f(f(x)^2y)=x^3y^3f(xy)$. Now, $P(xy,1) \Rightarrow f(f(xy)^2)=x^3y^3f(xy)=f(f(x)^2f(y)^2)$. Using Lemma 1, we get $f(xy)=f(x)f(y)$. $\Box$ Lemma 3 $f(x)=\frac{1}{x} \text{ } \forall x \in \mathbb{Q^+}$. PROOF: Let $g(x)=xf(x)$. Then using Lemma 2, we have $g(xy)=g(x)g(y)$. Substituting this in $P(x,y)$, we get $g(g(x))=g(x)^{\frac{5}{2}}$. Let $g^n(x)$ denote $g(g( \dots (g(x)) \dots))$, where the function $g$ is applied $n$ times. Using the above equality, one gets that $g^n(x)=g(x)^{\left(\frac{5}{2} \right)^{(n-1)}}$. As the LHS above is always a positive rational number, we get that the RHS must also be a positive rational number. Thus $g(x)$ must be $2^{(n-1)^{\text{th}}}$ power of some positive rational number $\forall n \in \mathbb{N}$ and $\forall x \in \mathbb{Q^+}$. As we vary $n$ over the set of natural numbers, one can easily see that $g$ must be equivalent to one for all positive rational values for the given condition to be met, i.e. $g \equiv 1$. Thus, we get the only solution as: $f(x)=\frac{1}{x} \text{ } \forall x \in \mathbb{Q^+}$ $\blacksquare$
07.10.2018 10:15
How does $g^n(x)=g(x)^{\left(\frac{5}{2} \right)^{(n-1)}}$ imply math_pi_rate wrote: $g(x)$ must be $2^{(n-1)^{\text{th}}}$ power of some positive rational number $\forall n \in \mathbb{N}$ and $\forall x \in \mathbb{Q^+}$. ?
07.10.2018 14:55
Cookie1234 wrote: How does $g^n(x)=g(x)^{\left(\frac{5}{2} \right)^{(n-1)}}$ imply math_pi_rate wrote: $g(x)$ must be $2^{(n-1)^{\text{th}}}$ power of some positive rational number $\forall n \in \mathbb{N}$ and $\forall x \in \mathbb{Q^+}$. ? As the LHS above is always a positive rational number, we get that the RHS must also be a positive rational number. This means that $g(x)^{\left(\frac{5}{2} \right)^{(n-1)}}$ is always a positive rational number. But this is true only when $g (x) $ is the $2^{(n-1)^{\text{th}}}$ power of some positive rational number $\forall n \in \mathbb{N}$ and $\forall x \in \mathbb{Q^+}$.
12.01.2019 09:27
Let $P(x,y)$ denote the assertion. From $P(x, a/x)$ we see that if $S =\operatorname{im} f$, then $s\in S\implies sx^3 \in S$ for all $x$ (let $f(a) = s$). From $P(x,1)$ we see that $f(f(x)^2) = x^3f(x)$, i.e. $x^3 = f(f(x)^2)/f(x)$ which implies that $x$ is injective. From $P(1,x)$ we see that $f(f(1)^2x) = f(x)$ so $f(1) = 1$. In particular, this means for each $x\in \mathbb Q^+$, there is a unique $y$, namely $y = f(x)^2/x$ (by taking $P(x, 1/x)$), such that $P(y) = x$. From $P(x,f(a)^2/(ax))$ we have \[ f\left( f(x)^2 \cdot \frac{f(a)^2}{ax}\right) = x^3 f\left( x\frac{f(a)^2}{ax}\right) = x^3a^3 = f\left( \frac{f(ax)^2}{ax}\right), \]so $f(x)f(a) = f(ax)$, i.e. $f$ is multiplicative, whereupon we get $f(f(x)^2) = x^3f(x)$. In particular, this means that $xf(x) = f(f(x)^2)/x^2 = (f(f(x))/x)^2$ is a square of a positive rational. Fix $x$ and let $xf(x) = a^2$. We claim $a = 1$. Note $f(x) = a^2/x$ and we get that $f(f(x))^2 = x^2a^2$, or $f(f(x)) = ax$, so $f(a^2/x) = ax$, or $f(a^2) = f(x)ax = a^3$. Now, call a positive rational murine if it satisfies $f(a^2) = a^3$. We claim the following: If $a$ is murine, then $f(a)^2 = a^3$, so $a$ is the square of a rational since $3\nu_p(a) = 2\nu_p(f(a))$ is even for all primes $p$, where $\nu_p$ is $p$-adic valuation. Furthermore, if $a = b^2$, we have \[ f(b^2)^2 = b^6\implies f(b)^2 = b^3 \]by multiplicity. In particular, if $a$ is murine, then $\sqrt a$ is a positive, murine rational. It follows that $a^{\frac 1{2^n}}$ is a positive, murine rational for all $n\in \mathbb N$, which forces $a = 1$ as desired. So, $f(x) = 1/x$ for all $x$.
16.04.2019 21:35
Denote the assertion as $P(x,y)$. $P(x,1)$ yields $f(f(x)^2)=x^3f(x)$. If we suppose $f(a)=f(b)$, then this equation implies $a^3=b^3\implies a=b$. So, $f$ is injective. $P(1,x)$ yields $f(f(1)^2x)=f(x)$, and by injectivity, we get that $f(1)=1$. Now, note that $f(xy)=\frac{f(f(x)^2y}{x^3}$. Applying this again, we get $f(f(x)^2y)=\frac{f(f(x)^2f(y)^2)}{y^3}$. So, $f(xy)=\frac{f(f(x)^2f(y)^2)}{x^3y^3}$. Now, by injectivity, we get that $f(x)^2f(y)^2=f(w)^2f(z)^2$ for all $xy=wz$. So, we can same that $f(xy)=f(xy)f(1)=f(x)f(y)$, and $f(x)$ is multiplicative. So, our original assertion states that $f(f(x))^2=x^3f(x)$. Consider a prime $p$, and suppose that $f(p)=\frac{g(p)}{p}$. Then, when we plug in, we get that $g(p)^3=f(g(p))^2$. As both sides are rational numbers, we get that $g(p)=g_2(p)^2$ for some rational $g_2(p)$. After substituting again, we get that $g_2(p)^3=f(g_2(p))^2$, so $g_2$ is also a square. Continuing downwards, we find that $g(p)$ is a perfect $2^k$ power for arbitrarily large $k$, so we must have $g(p)=1$, and $f(p)=\frac{1}{p}$. Now, of course, by multiplicativity, $f(n)=\frac{1}{n}$ for all $n$, and this clearly works. So, this is the only solution.
10.12.2019 21:46
We will find $1/f$ by solving $f(xy)=x^3f(yf(x)^{-2}).$ Denote this relation by $P(x,y)$. Fact 1.$f(xf(x)^2)=x^3.$
Fact 2. If $f(x)=f(xc) \ \forall x$, then $c=1$.
Fact 3. $f(1)=1.$
Fact 4. $f$ is injective.
Fact 5. $f(xy)=f(x)f(y).$
Thus $f(x)f(f(x))^2=x^3$ by facts 1 and 5. Hence $f(yf(x))^2=f(f(x))^2f(y)^2=\frac{x^3}{f(x)}f(y)^2.$ Then $\frac{f(x)}{x}=(\frac{f(yf(x))}{xf(y)})^{-2}$. Putting $y=x$ and $g(x)=xf(x)$ we have $$\frac{f(x)}{x}=(\frac{f(g(x))}{g(x)})^{-2}=(\frac{f(g(g(x)))}{g((g(x))})^{(-2)^2}=(\frac{f(g^t(x))}{g^t(x)})^{(-2)^t}$$So $(\frac{f(x)}{x})^{1/(-2)^t}$ is rational for all $t \in \mathbb{N}$, then clearly $f(x)=x$. Therefore $1/x$ is the only solution to the original equation, which clearly works.
19.03.2020 17:02
30.04.2020 07:38
Claim: $f$ is injective and $f(1)=1$. Proof: $P(x,1)$ gives $f(f(x)^2)=x^2f(x)$, which is something with only $f(x)$ and an uwrapped $x$. Hence if $f(a)=f(b)$ then clearly $a=b$. So $f$ is injective. Now, $P(1,1)$ gives $f(f(1)^2) = f(1)$, so $f(1)^2=1$, and hence $f(1)=1$. $\square$ Claim: $f(xy)=f(x)f(y)$. Proof: $P(x,y)$ gives $f(f(x)^2y)=x^3f(xy)$. We want to get the above LHS in another way from the FE: $P(y,f(x)^2)$ gives $f(f(y)^2f(x)^2)=y^3f(yf(x)^2)$. So $f(yf(x)^2) = \tfrac{1}{y^3} f(f(y)^2f(x)^2)$. Therefore, \[ \tfrac{1}{y^3}f(f(y)^2f(x)^2) = x^3f(xy) \implies f(f(y)^2f(x)^2) = x^3y^3f(xy). \]We want to get rid of the $x^3y^3$, so $P(xy,1)$ gives $f(f(xy)^2)=x^3y^3f(xy)$. Therefore, \[ f(f(y)^2f(x)^2) = f(f(xy)^2) \implies f(x)^2f(y)^2 = f(xy)^2 \implies f(x)f(y)=f(xy), \]as desired. $\square$ Now, using multiplicativity to split up, the original FE becomes $f(f(x))^2f(y)=x^3f(x)f(y)$, i.e. $f(f(x))^2 = x^3 f(x)$. Let $g(x)=xf(x)$. Then \[ f(f(x)) = \frac{g(f(x))}{f(x)} = \frac{g(g(x)/x)}{g(x)/x} = \frac{g(g(x)) g(1/x)}{g(x)/x} = \frac{xg(g(x))}{g(x)^2}\]where we used multiplicativity multiple times; note that $g$ is also multiplicative since $f$ is. So \[ f(f(x)) = x^{3/2} \sqrt{f(x)} \implies \frac{xg(g(x))}{g(x)^2} = x^{3/2} \sqrt{g(x)/x} \implies g(g(x)) = g(x)^{5/2}.\]Therefore, $g^{k+1}(x) = g(x)^{5^k/2^k}$, for any $k\ge 1$. Unless $g(x)=1$, this will eventually become a non-rational. Therefore, $g(x)=1$ for all $x$, and hence $xf(x)=1$, so $f(x)=1/x$ for all $x$. It clearly works.
30.04.2020 11:04
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29.08.2020 20:34
We claim $f(x)=1/x$ is the only solution. It's trivial to see that this works, so now suppose $f$ is a function satisfying the FE. Let $P(x,y)$ denote the given FE. Note that \[P(1,y)\implies f(f(1)^2 y)=f(y),\]and \[P(f(1)^2 x,y)\implies f(f(x)^2 y) = f(1)^6 x^3f(xy),\]so $f(1)^6=1$, so $f(1)=1$. We claim that $f$ is injective. We see that \[P(x,1)\implies f(f(x)^2) = x^3f(x).\]If $f(a)=f(b)$, then the above equation with $a$ and $b$ implies $a=b$, so $f$ is injective. Now, \[P(x,1/x)\implies f\left(\frac{f(x)^2}{x}\right) = x^3.\]This implies $f$ is surjective as well, so $f$ is a bijection. This also implies that $x\mapsto f(x)^2/x$ is also a bijection. Now, \[P(x,y/x)\implies f\left(\frac{f(x)^2}{x}y\right) = x^3 f(y) = f\left(\frac{f(x)^2}{x}\right)f(y),\]and since $x\mapsto f(x)^2/x$ is a bijection, we see that \[f(xy)=f(x)f(y)\]for all $x$ and $y$. The original FE now reduces to $f(f(x))^2 = x^3f(x)$. It is easy to see that this is equivalent to \[\frac{f(xf(x))^2}{(xf(x))^3}=1,\]so it suffices to show that $x\mapsto f(x)^2/x^3$ is injective, which would imply $xf(x)=1$, or that $f(x)=1/x$, as desired. Suppose $f(a)^2/a^3=f(b)^2/b^3$. Then, we get $f(a/b)^2=(a/b)^3$, so letting $\alpha=a/b$, we see that $\alpha$ must be a square of a rational, say $\alpha=\beta^2$. Then, the equation becomes $f(\beta)^2=\beta^3$, so $\beta$ must also be a square of a rational, and so on. This is not possible unless $\alpha=1$, so $x\mapsto f(x)^2/x^3$ is injective, so we're done. Remark: We'll provide some motivational remarks for the solution after we get that $f$ is multiplicative. Indeed, one can view $f$ as a linear map $V\to V$, where $V$ is a the $\mathbb{Z}$-module of vectors $(x_1,x_2,\ldots)$ where the $x_i$ are integers, and all but finitely many of them are $0$. This represents $\mathbb{Q}$ through prime factorization. Suppose this linear map is called $T$. Then, the equation $f(f(x))^2 = x^3f(x)$ becomes \[T^2 = T+3\mathrm{id},\]which factors as \[(2T-3\mathrm{id})(T+\mathrm{id}) = 0.\]It suffices now to show that $2T-3\mathrm{id}$ has trivial kernel. However, linear algebraically, this has no right to be true, and we can actually construct linear maps $T$ (and thus supposedly functions $f$) for which this fails. The key here is that $T$ is a map between $\mathbb{Z}$-modules rather than vector spaces, so we have an extra divisibility structure. We now see that any element of the kernel of $2T-3\mathrm{id}$ must have all even entries, but that means we can divide it by $2$, and it will still remain in the kernel, which is a contradiction unless the element is all $0$s. This becomes the repeated square root argument.
31.08.2020 16:25
For imo, what books should I study for Functional Equation?
24.11.2020 13:31
IMO 2010 SL A5 wrote: Denote by $\mathbb{Q}^+$ the set of all positive rational numbers. Determine all functions $f : \mathbb{Q}^+ \mapsto \mathbb{Q}^+$ which satisfy the following equation for all $x, y \in \mathbb{Q}^+:$ \[f\left( f(x)^2y \right) = x^3 f(xy).\] Proposed by Thomas Huber, Switzerland As usual, let $P(x,y)$ be the given FE. We start with some Claims: Claim 1: $f$ is injective and $f(1)=1$. Proof: To prove injectivity, just consider $P(x,1)$. In addition, $P(1,1) \Rightarrow f(f^2(1))=f(1)$, which implies $f(1)=1$ since $f(1)>0$ and $f$ is injective $\blacksquare$. Claim 2: $f(x^6)=f^6(x)$ Proof: $P(x,\dfrac{1}{x})$ implies $f(\dfrac{f^2(x)}{x})=x^3, \,\, (*)$ and $P(x,1)$ implies $f(f^2(x))=x^3f(x), \,\, (**)$, so by taking $x \rightarrow f^2(x)$ at $(*)$ and using $(**)$, we obtain the desired $\blacksquare$. Claim 3: $f(f(x))=\sqrt{x^3f(x)}$. Proof: Consider $P(x,\frac{f^2(x)}{x^2})$, to obtain $$f(\frac{f^4(x)}{x^2})=x^3f(\dfrac{f^2(x)}{x})=x^6,$$so by taking $x \rightarrow f(x)$ at the latter and using injectivity, we obtain $$\frac{f^4(f(x))}{f(x)^2}=x^6 \Rightarrow f(f(x))=\sqrt{x^3f(x)},$$as desired $\blacksquare$. To the problem, by the last Claim's result we have that $x^3f(x)$, or equivalently $xf(x)$, must be the square of a rational number, for all $x$. Put $x \rightarrow f(x)$ at $f(f(x))=\sqrt{x^3f(x)}$, to obtain that $f(x)^3f(f(x))$ is the square of a rational as well. But, $$f(x)^3f(f(x))=f(x)^3\sqrt{x^3f(x)}=\sqrt[4]{f(x)^7x^3},$$so $f(x)^7x^3$, or equivalently $(f(x)x)^7$, must be a perfect fourth power for all $x$. Since $\gcd(4,7)=1$, we actually have that $xf(x)$ is a perfect fourth power. We easily see a pattern. In order to formalize it, we do the following: We define each iteration as a step. Let the powers of $f(x)$ and $x$ at the $i_{\rm th}$ step be $a_i$ and $b_i$, while the exponent is $1/2^i$. So, in the first step, $a_1=1,b_1=3$ and the exponent is $1/2$ - namely, the square root. At the $(i+1)_{\rm th}$ step, the expression $\sqrt[2^i]{f(x)^{a_i}x^{b_i}}$ transforms to $\sqrt[2^i]{f(f(x))^{a_i}f(x)^{b_i}}$, which after some algebraic manipulations turns out to be $\sqrt[2^{i+1}]{f(x)^{a_i+2b_i}x^{3a_i}}$. That is, $a_{i+1}=a_i+2b_i$ and $b_{i+1}=3a_i$ for all $i$. In addition, if we define $c_i=b_i-a_i$ for each $i$, then $c_{i+1}=b_{i+1}-a_{a+1}=2(a_i-b_i)=2c_i$, and since $c_1=b_1-a_1=2$ we have that $c_i=2^i$ for all $i$. Now, note that $a_{i+1}=a_i+2b_i \equiv a_i \pmod 2$, hence all $a_i$ are odd. Finally, we have $$\sqrt[2^i]{f(x)^{a_i}x^{b_i}}=\sqrt[2^i]{(f(x)x)^{a_i}} \cdot x,$$since $b_i-a_i=2^i$, and since $a_i$ is odd, $f(x)x$ is a perfect ${2^i}_{\rm th}$ power, for all $i$. But this readily implies that $f(x)x=1$, therefore $f(x)=\frac{1}{x}$, which evidently satisfies, so it is our only solution.
05.03.2021 09:41
orl wrote: Denote by $\mathbb{Q}^+$ the set of all positive rational numbers. Determine all functions $f : \mathbb{Q}^+ \mapsto \mathbb{Q}^+$ which satisfy the following equation for all $x, y \in \mathbb{Q}^+:$ \[f\left( f(x)^2y \right) = x^3 f(xy).\] Proposed by Thomas Huber, Switzerland Ez. Let $P(x, y)$ be the assertion. We claim that $\boxed{f(x) = \frac{1}{x}}$ is the only solution to the given functional equation, which indeed works so it suffices to show that this is the only solution. If $f(a) = f(b)$, then $P(a, 1) \implies f(f(a)^2) = a^3f(a) \dots (1) $ and similarly $P(b, 1) \implies f(f(b)^2) = b^3f(b) \dots (2)$ and since $f(a) = f(b) = c$, we get that $f(c^2) = a^3c = b^3c$ implying that $a = b$ or $f$ is injective. Now, $P(1, 1) \implies f(f(1)^2) = f(1) \implies f(1)^2 = f(1)$ or $\boxed{f(1) = 1}$ Now, $P(a, f(b)^2) \implies f(f(a)^2f(b)^2) = a^3f(af(b)^2) = (ab)^3f(ab) \dots (3)$ and $P(ab, 1) \implies f(f(ab)^2) = (ab)^3f(ab) \dots (4)$, so comparing $(3)$ and $(4)$ and since $f$ is injective, we get that $f(a)^2f(b)^2 = f(ab)^2$ or $f(a)f(b) =f(ab)$ implying that $f$ is multiplicative. Now, if $g(x) = xf(x)$, we get that $g^n(x) = g(x)^{\frac{5}{2}^{n-1}}$ and so if $g(x) \neq 1$, then $g^n(x)$ for all sufficiently large enough $n$ will not be rational, a contradiction since domain of $g$ must also be rational. Therefore, $g(x) = 1$ for all positive rationals $x$ implying that $f(x) = \frac{1}{x}$ for all positive rationals $x$ as desired
24.07.2021 04:30
Let $P(x,y)$ be the given assertion, and for any function $h$, $h^n(x)=\underbrace{h(h(\ldots h(}_{n\text{ times}}x)\ldots))$. $P(1,x)\Rightarrow f(xf(1)^2)=f(x)$ $P(xf(1)^2,y)-P(x,y)\Rightarrow f(1)=1$ $P(x,1)\Rightarrow f(f(x)^2)=x^3f(x)\Rightarrow f$ is injective. $P(x,f(y)^2)-P(y,x)-P(xy,1)\Rightarrow f(f(x)^2f(y)^2)=f(f(xy)^2)\Rightarrow f(xy)=f(x)f(y)$ Let $g(x)=xf(x)$, so that $g(xy)=g(x)g(y)$, $g$ is injective, and $g(1)=1$. $P(x,1)\Rightarrow g(g(x))=g(x)^{5/2}$, so $g^{n+1}(x)=g(x)^{\left(\frac52\right)^n}$ for all $n\in\mathbb N_0$ by simple induction. However, for sufficiently large $n$, $g(x)^{\left(\frac52\right)^n}$ is not rational unless $g(x)=1$. Then $\boxed{f(x)=\frac1x}$, which works.
21.01.2022 19:05
it is not difficult to find $f(xy)=f(x)f(y)$ from a simple value. $$f(f(x)) = x ^ \frac{3}{2} \cdot f(x) ^ \frac{1}{2}$$It is possible to prove the rationality of $f(x)^\frac{1}{2^n} \cdot x^\frac{1}{2^n}$ from induction. For an arbitrary $n$ natural number, then $xf(x)=1$ Answer:$ \boxed {f(x)=\frac {1}{x}}$
19.05.2022 20:50
Let $P(x,y)$ denote the given assertion. $P(x,1)\implies f(f(x)^2)=x^3f(x).$ So $f$ is injective. Indeed $f(1)=1,$ and comparing $P(x,f(y)^2)$ and $P(xy,1)$ yields that $$f(xy)=f(x)f(y).$$$P(x,y)\implies f(f(x))=\sqrt{x^3f(x)}.$ Now letting $g(x)=xf(x)$ yields $$g(g(x))=g(x)^{5/2}.$$By induction, $$g^{n+1}(x)=g(x)^{(5/2)^n},$$where $n$ is natural. Notice that $g(x)\neq 1$ is absurd since $g(x)^{(5/2)^n}$ is irrational for large $n.$ Thus $f(x)=1/x,$ which fits.
05.06.2022 18:11
We claim $f(x) = \frac{1}{x}$ is the only solution. It works since \begin{align*} f(f(x)^2y) = \frac{x^2}{y} = x^3f(xy). \end{align*} Let $P(x, y)$ denote the assertion that $f(f(x)^2y) = x^3f(xy)$. Claim 1: $f$ is injective. Proof: $P(x, 1)$ gives $f(f(x)^2) = x^3f(x)$. Now $f(\alpha) = f(\beta) \Rightarrow \alpha^3 = \beta^3 \Rightarrow \alpha = \beta$ in positive rationals. $\square$ Claim 2: $f(1) = 1$. Proof: $P(1, x)$ gives $f(f(1)^2x) = f(x)$ so by injectivity, $f(1)^2 = 1 \Rightarrow f(1) = 1$. $\square$ Claim 3: $f$ is multiplicative. Proof: $P(x, f(y)^2)$ gives $f(f(x)^2f(y)^2) = x^3f(xf(y)^2) = x^3y^3f(xy)$. $P(xy, 1)$ gives $f(f(xy)^2) = x^3y^3f(xy)$. Hence by injectivity, $f(x)^2f(y)^2 = f(xy)^2$, and so working in positive rationals, $f(xy) = f(x)f(y)$. $\square$ Now we reduce $P(x)$ to the following equivalent assertion: $f(f(x))^2 = x^3f(x)$. Claim 4: $[f^n(x)]^{2^{n-1}} = x^{a_n}f(x)^{b_n}$, where $|a_n - b_n| = 2^{n-1}$ and both $a_n, b_n$ are odd. Proof: We induct on $n$. Base case: $n = 2$ done by $P(x)$. Inductive step: Assume holds for $n-1$. \begin{align*} [f^n(x)]^{2^{n-1}} &= ([f^n(x)]^{2^{n-2}})^2 \\ &= ([f^{n-1}(f(x))]^{2^{n-2}})^2 \\ &= (f(x)^{a_{n-1}}f(f(x))^{b_{n-1}})^2 \\ &= f(x)^{2a_{n-1}} (f(f(x))^2)^{b_{n-1}} \\ &= f(x)^{2a_{n-1}} x^{3b_{n-1}} f(x)^{b_{n-1}} \\ &= x^{3b_{n-1}} f(x)^{2a_{n-1} + b_{n-1}}. \end{align*} Now $|a_n - b_n| = |3b_{n-1} - 2a_{n-1} - b_{n-1}| = 2|b_{n-1} - a_{n-1}| = 2 \cdot 2^{n-2} = 2^{n-1}$, and $a_n = 3b_{n-1}$ is odd, $b_n \equiv b_{n-1} \pmod{2}$ is also odd. $\square$ Claim 5: $f(x) = \frac{1}{x}$. Proof: Consider any prime $p$. \begin{align*} v_p([f^n(x)]^{2^{n-1}}) &= v_p(x^{a_n}) + v_p(f(x)^{b_n}) \\ \Rightarrow 2^{n-1} v_p(f^n(x)) &= a_n v_p(x) + b_n v_p(f(x)) \\ \Rightarrow a_n v_p(x) + b_n v_p(f(x)) &\equiv 0 \pmod {2^{n-1}} \\ \Rightarrow a_n v_p(x) &\equiv - b_n v_p(f(x)) \pmod{2^{n-1}} \\ \Rightarrow \frac{a_n}{b_n} v_p(f(x)) &\equiv - v_p(x) \pmod{2^{n-1}} \\ \Rightarrow v_p(f(x)) &\equiv -v_p(x) \pmod{2^{n-1}} \end{align*}where $\frac{a_n}{b_n} \equiv 1 \pmod{2^{{n-1}}}$ follows from the conditions on $a_n, b_n$ in Claim 4. Now note that this holds for all $n$ large enough, and so $v_pf(x) = -v_p(x)$ for all primes $p$. Hence $f(x) = \frac{1}{x}$. $\blacksquare$
10.06.2022 12:43
06.07.2022 04:10
We claim $f(x) = \frac{1}{x}$ is the only solution. Note that it works. Let $P(x,y)$ be the assertion \[f(f(x)^2y) = x^3f(xy)\]\[P(1,1): f(f(1)^2) = f(1)\]\[P(f(1)^2,1): f(1) = f(f(f(1)^2)^2) = f(1)^6f(f(1)^2) = f(1)^7 \therefore f(1) = 1\]\[P(x,1) :f(f(x)^2) = x^3f(x)\]\[P(f(x)^2,\frac{y}{f(x)^2}): f(f(f(x)^2)^2\frac{y}{f(x)^2}) = f(x)^6f(y)\]\[\implies f(x^6y) = f(x)^6f(y)\]\[\implies f(x^6) = f(x)^6\]\[\implies f(xy)^6 = f(x)^6f(y)^6\]\[\therefore f(xy) = f(x)f(y)\] Let $g(x) = x^2$ \[P(x,1):g^2(x) =g(x)x^6 \]Solving the recurrence for $g^n(x)$, and using $f(x^2) = f(x)^2$, \[f(x)^{2^n} = g^n(x) = g(x)^{\frac{3^n-(-2)^n}{5}}x^{\frac{2.3^n+3.(-2)^n}{5}}\]Taking $\nu_p's$ and looking $\pmod 2^n$: \[\frac{2.3^n}{5}(\nu_p(f(x))+\nu_p(x))\equiv 0 \pmod 2^n.\]Taking $n$ large, we get that $\nu_p(f(x))+\nu_p(x) = 0$ for all primes $p$, so $f(x) = \frac{1}{x}$
07.07.2022 17:46
Fun (...ctional equation) problem The only solution is $f(x) \equiv \frac{1}{x}$, which works. Let $P(x,y)$ denote the assertion that $f(f(x)^2y)=x^3f(xy)$. By $P\left(x,\frac{y}{x}\right)$, we have \[f\left(\frac{f(x)^2}{x}y\right)=x^3f(y).\]By $P\left(\frac{f(a)^2}{a}x,y\right)$, we have \[f\left(f\left(\frac{f(a)^2}{a}x\right)^2y\right)=\frac{f(a)^6}{a^3}x^3f\left(\frac{f(a)^2}{a}xy\right) \Rightarrow f(a^6f(x)^2y)=\frac{f(a)^6}{a^3}x^3 \cdot a^3f(xy)=f(a)^6x^3f(xy)=f(a)^6f(f(x)^2y).\]Since $f(x)^2y$ can be any rational number, we have $f(a^6b)=f(a)^6f(b)$ for all $a,b \in \mathbb{Q}^+$. Plug in $(a,b)=(1,1)$ to get $f(1)=1$, and plug in $(a,1)$ to get $f(a^6)=f(a)^6$. Then, plug in $b^6$ into $b$ to get \[f(a^6b^6)=f(a)^6f(b^6) \Rightarrow f(ab)^6=f(a)^6f(b)^6 \Rightarrow f(ab)=f(a)f(b).\]Therefore, the original equation becomes \[f(f(x)^2y)=x^3f(xy) \Rightarrow f(f(x))^2f(y)=x^3f(x)f(y) \Rightarrow f(f(x))^2=x^3f(x).\] Fix $x$, and suppose that $f(x)=\frac{k}{x}$. Then, we have \[f\left(\frac{k}{x}\right)^2=x^3 \cdot \frac{k}{x} \Rightarrow \frac{f(k)^2}{f(x)^2}=\frac{f(k)^2}{k^2} \cdot x^2=kx^2 \Rightarrow f(k)^2=k^3.\]If $k \ne 1$, then let $m$ be the largest integer such that $\sqrt[2^m]{k}$ is rational, and let $k_0=\sqrt[2^m]{k}$. Notice that $k_0$ is not the square of a rational number. We see that \[f(k_0^{2^m})^2=f(k_0)^{2^{m+1}}=k_0^{3 \cdot 2^m} \Rightarrow f(k_0)^2=k_0^3.\]However, $k_0^3$ is not a square of a rational number, so $k=1$ and we have $f(x) \equiv \frac{1}{x}$ for all $x \in \mathbb{Q}^+$.
01.12.2022 16:51
21.02.2023 02:36
Let $P(x,y)$ denote the assertion. Then, $P(x,1)$ gives $f(f(x)^2)=x^3f(x)$. If $f(x)$ is constant then $f(f(x)^2)$ and $f(x)$ are constant so $x$ is constant. Thus, $f$ is injective. Thus, $P(1,1)$ gives $f(1)=1$. Now, $P(x,f(y)^2)$ gives $f(f(x)^2f(y)^2=x^3f(xf(y)^2))=x^3y^3f(xy)$ but $P(xy,1)$ gives $f(f(xy)^2)=x^3y^3f(xy)$ so $f(x)f(y)=f(xy)$. This reduces the equation to $Q:f(f(x))^2=x^3f(x)$ so let $g(x)=xf(x)$ which is also multiplicative. Then, $Q$ becomes $g(g(x))=g(x)^{2.5}$. Then, $g(g(g(x)))=g(g(x))^{2.5}=g(x)^{6.25}$ and iteratively we get that $g^{n+1}(x)=g(x)^{{2.5}^n}$. Unless $g$ is $1$, this is irrational, which is absurd. Thus, $f(x)=\tfrac1x$.
01.06.2023 04:59
Airplane solve :skulls: The only solution is $\boxed{f(x) = \frac{1}{x}}$. To see this works, notice the LHS expands to $\frac{1}{\left(\tfrac{1}{x} \right)^2 y} = \frac{x^2}{y}$ and the RHS expands to $\frac{x^3}{xy} = \frac{x^2}{y} $ also. Now we prove it is the only solution. Let $P(x,y)$ denote the given assertion. $P(x,1): f(f(x)^2) = x^3 f(x)$. $P(1, x): f(x f(1)^2) = f(x)$. Let $c = f(1)^2$. Claim: There does not exist $c\ne 1$ such that $f(x) = f(cx)$ for all $x$. Proof: Suppose there existed such a $c$. Comparing $P(x,y)$ with $P(cx,y)$ we get $x^3 = (cx)^3$, so $c = 1$, contradiction. $\square$ Therefore $P(1,x)$ combined with the claim gives that $f(1)^2 = 1\implies f(1) = 1$. Claim: $f$ is injective. Proof: Suppose $f(a) = f(b)$. Then $P(a,1)$ compared with $P(b,1)$ implies $a^3 = b^3$, so $a = b$, as desired. $\square$ $P\left(x, \frac{1}{f(x)^2} \right): x^3 \cdot f\left( \frac{x}{f(x)^2}\right) = 1$, so $f\left(\frac{x}{f(x)^2} \right) = \frac{1}{x^3}$. $P\left(\frac{1}{x}, x\right): f\left(f\left(\frac{1}{x} \right)^2 \cdot x\right) = \frac{1}{x^3}$. Combining these two equations with injectivity, we deduce \[ f\left(\frac{1}{x} \right)^2 \cdot x = \frac{x}{f(x)^2} \implies f(x) \cdot f\left( \frac{1}{x} \right) = 1\] $P\left(x, \frac{1}{x} \right): f\left( \frac{f(x)^2}{x} \right) = x^3$. Claim: Let $S$ be the set of all positive rationals expressible in the form $\frac{f(x)^2}{x}$ for some positive rational $x$. Then for any $s\in S$ and $k\in $, we have $f(sk) = f(s)f(k)$. Proof: Using $P\left(x, \frac{1}{x} \right)$, we can rewrite the FE as \[ f(f(x)^2 y) = f\left(\frac{f(x)^2}{x} \right) f(xy)\]Subbing $y = \frac{k}{x}$ for any $k\in \mathbb{Q}^{+}$ in the equation gives \[f\left(\frac{kf(x)^2}{x} \right) = f\left( \frac{f(x)^2}{x} \right) f(k)\]Subbing $\frac{f(x)^2}{x}$ for $s$ (any $s\in S$ can be expressed in this form), we get $f(sk) = f(s) f(k)$, as desired. $\square$ $P(xy,1)$ compared with the original FE gives \[f(f(xy)^2) = (xy)^3 f(xy) = f(f(x)^2 y ) \cdot y^3 = f(f(x)^2 y) f\left( \frac{f(y)^2}{y} \right) \] Now substituting $s = \frac{f(y)^2}{y}$ and $k = f(x)^2 y $, we get $ f(f(x)^2 y) f\left( \frac{f(y)^2}{y} \right) = f(f(x)^2 f(y)^2)$. Now, this must be equal to $f(f(xy)^2)$, so using injectivity, \[f(x)^2 f(y)^2 = f(xy)^2 \implies (f(x)f(y))^2 = f(xy)^2 \implies f(x)f(y) = f(xy)\forall x,y\in \mathbb{Q}^{+}\] We may rewrite the equation as $f(f(x)^2) = x^3 f(x)$, so $\frac{f(f(x)^2)}{f(x)} = x^3$, and that is equivalent to \[ f\left( \frac{f(x)^2}{x} \right) = x^3\]In other words, we must find all functions over $\mathbb{Q}^{+}$ satisfying the above equation and $f(xy) = f(x) f(y)$ for all $x,y \in \mathbb{Q}^{+}$. Let $g(x) = xf(x)$. We have \[ f\left( \frac{g(x)^2}{x^3} \right) = x^3 \implies g\left( \frac{g(x)^2}{x^3} \right) = g(x)^2 \]and $g(xy) = g(x) g(y) $. We then find that $g\left( \frac{g(x)^2}{x^3} \right) \cdot g(x) = g(x)^3$, so \[ g\left( \frac{g(x)^2}{x^2} \right) = g(x)^3 \implies g(g(x)^2) = g(x)^5 = g(g(x))^2\]Thus, $g(x)^5$ is a perfect square, so $g(x)$ must be a perfect square for all $x\in \mathbb{Q}^{+}$. Let $g(x) = g_1(x)^2$, so that $g_1(x)$ inputs and outputs only positive rational numbers. Then denote $g_2(x) = \sqrt{g_1(x)}$, and $g_k(x) = \sqrt{g_{k-1}(x)}$ for any positive integer $k\ge 2$. Claim: $g_n\left (g_n(x)^{2^{n+1}}\right ) = g_n(x)^5$ for each positive integer $n$ and positive rational $x$. Proof: We induct on $n$. First we prove the base case. Notice that using $g(g(x)^2) = g(x)^5$, we have $g_1(g_1(x)^4) ^2 = g(x)^{10}$, so $g_1(g_1(x)^4) = g_1(x)^5$, so $n=1$ is true. Suppose the result was true for everything up to $t-1$. Then using the fact that $g_{t-1} \left(g_{t-1}(x)^{2^t}\right) = g_{t-1}(x)^5$, we get \[g_t \left( (g_t(x) ^2)^{2^t} \right) ^2 =(g_t(x)^2)^5\implies g_t\left(g_t(x)^{2^{t+1}} \right) ^2 = g_t(x)^{10},\]and then taking the square root of both sides gives the desired result, thus completing the induction. $\square$ Therefore $g_n(x)^5 =(g_n(g_n(x))^{2^{n + 1}} = g_n(x)^5$, so $g_n(x)^5$ is a perfect square, which implies $g_n(x) $ is a perfect square for each $n$. Therefore $g_n$ also must output positive rationals for each positive integer $n$. Also we note that $g(x) = g_n(x) ^{2^n}$, which means $g(x)$ is a perfect $2^n$th power for each positive integer $n$, thus $g(x) = 1$ for all $x$. So \[f(x) =\frac{g(x)}{x} = \frac{1}{x} \forall x\in \mathbb{Q}^{+},\]as desired.
04.06.2023 15:26
Solved with Mirhabib Let $P(x,y)$ denote the assertion $$P(x,1) \implies f(f(x)^2)=x^3f(x) \qquad (1)$$If $f(a)=f(b)$ replacing $x$ with $a$ and $b$ in $(1)$ gives us $a^3=b^3 \implies a=b \implies f \text{ is injective }$ $P(1,1) \implies f(1)=1$. So $f(\frac{f(x)^2}{x})=x^3$ and $f(\frac{x}{f(x)^2})=\frac{1}{x^3}.$ $P(\frac{1}{y},y)$ gives $f(f(\frac{1}{y})^2y)=\frac{1}{y^3}$. Hence $$f(\frac{x}{f(x)^2})=f(f(\frac{1}{x})^2x) \implies \frac{x}{f(x)^2}=f(\frac{1}{x})^2x \implies f(x)f(\frac{1}{x})=1$$$$\text{ In 1 replace x with } \frac{f(x)^2}{x} \implies f(x^6)=f(x)^6 \qquad (2)$$$$P(f(x)^2 , \frac{y^6}{f(x)^2}) \implies f(x^6y^6)=f(x)^6f(y^6) \qquad (3)$$By $2$ and $3$ we get $f(xy)=f(x)f(y) \implies f$ is multiplicative.Let $g(x)=xf(x)$. Note that $g$ is also multiplicative and $g(x)g(\frac{1}{x})=1$.So $$f( f(x)^2y ) = x^3 f(xy) \iff f(f(x))^2=x^3f(x) \iff g(g(x))^2=g(x)^5$$By induction we get $$g^n(x)=g(x)^{{2.5}^{n-1}} $$Unless $g \equiv 1$, this will eventually become a non-rational. Therefore, $g \equiv 1$ and $\boxed{f(x)=\frac{1}{x}}$
24.08.2023 08:34
Just a bunch of mindless substitutions, though they were indeed pretty motivated... I hope FE dies sometime soon (figuratively). Notice $$(x,1):f(f(x)^2)=x^3f(x)\quad (1)\implies\forall f(a)=f(b):x=a,b\implies a=b;(1,1)\implies f(1)=1;(xy,1),(y,x),(x,f(y)^2):f(f(xy)^2)=x^3y^3f(xy)=x^3f(xf(y)^2)=f((f(x)f(y))^2)\implies f(x)f(y)=f(xy).$$Setting $$g(x)=xf(x)\in(1)\implies x^2g(x)=f(g(x)^2/x^2)=\frac{g(\frac{g(x)^2}{x^2})}{\frac{g(x)^2}{x^2}}\implies g(x^3)=g(\frac{g(x)^2}{x^2})\implies g(x)=x^{\frac52}\implies g^n(x)=x^{{\frac52}^n};$$in particular, this is not rational for sufficiently large n since $x=\frac pq$, to be rational q has to be rational but squarerooting it an arbitrary amount of times will make q irrational. So $g(x)=1\implies f(x)=\frac1x$, as desired. $\blacksquare$ wow 2010 a5 and no feeling whatsoever lol
01.02.2024 13:40
Let $P(x, y)$ be this assertion Claim: $f$ is injective Proof: Consider $P(x, 1)$. Then, $f(f(x)^2) = x^3f(x)$. Now, if $f(a) = f(b)$, then $f(a)^2 = f(b)^2$, so $a^3f(a) = f(f(a)^2) = f(f(b)^2) = b^3f(b)$ and $f(a) = f(b)$ so $a^3 = b^3$ and thus $a = b$. Now, take $P(1, y)$, and we get that $f(f(1)^2y) = f(y)$ so $f(1)^2y = y$ and thus $f(1) = 1$. Claim: $f(x^2) = f(x)^2$, $f\left(\frac{1}{x}\right) = \frac{1}{f(x)}$ and $f(x^3) = f(x)^3$ Proof: $P\left(x, \frac{1}{x}\right)$ implies that $f\left(\frac{f(x)^2}{x}\right) = x^3$, so putting $y = \frac{f(x)^2}{x^2}$, we get that $f\left(\frac{f(x)^4}{x^2}\right) = x^6$. But now by $P\left(x^2, \frac{1}{x^2}\right)$ we get $f\left(\frac{f(x^2)^2}{x^2}\right) = x^6 = f\left(\frac{f(x)^4}{x^2}\right)$, so by injectivity we get that $f(x)^2 = f(x^2)$. Now, we have that $f\left(\frac{f(x)^2}{x}\right) = x^3$ and in this equation replacing $x$ with $\frac{1}{x}$ and then taking $P\left(x, \frac{1}{f(x)^2}\right)$ we get that $f\left(\frac{x}{f(x)^2}\right) = \frac{1}{x^3}$, so by injectivity we will get that $f\left(\frac{1}{x}\right) = \frac{1}{f(x)}$ By $P(x, x^2)$ we get that $f(xf(x))^2 = x^3f(x^3)$. Now, by $P(x, f(x))$ we see that $f(f(x)^3) = x^3f(xf(x)$, and then squaring both sides we see that $f(f(x)^6) = x^9f(x^3)$, then by $P(x^3, 1)$ we see that $f(f(x^3)^2) = x^9f(x^3)$, so $f(x)^6 = f(x^3)^2$ so $f(x^3) = x^3$ Now $f(xf(x))^2 = x^3f(x^3) = (xf(x))^3$. Now let $S$ be the set of numbers that can be represented as $xf(x)$ for some $x$. Then, $t \in S$ implies $f(t)^2 = t^3$ implying that $tf(t) = t^{\frac{5}{2}} \in S$. But note that eventually $t^{\frac{5}{2}}$ is not defined in rationals. So, $S = {1}$ and thus $f(x) = \frac{1}{x}$ for all $x$ which clearly works.
07.06.2024 05:01
Solved with ihatemath123, GrantStar, and golue3120.
02.12.2024 06:53
Putting $x=1$ we get $f(f(1)^2y)=f(y)$. Putting $x=f(1)^2t$ we get \begin{align*} f(f(f(1)^2t)^2y)=f(1)^6t^3f(f(1)^2ty) & \implies f(f(t)^2y)=f(1)^6t^3f(ty) \\ & \implies t^3f(ty)=f(1)^6t^3f(ty) \\ & \implies f(1)=1.\end{align*} Claim: $f$ is multiplicative. Proof. Setting $y=1$ we get $f(f(x)^2)=x^3f(x)$. Taking $x=f(t)^2$ we get \begin{align*}f(f(f(t)^2)^2y) = f(t)^6f(f(t)^2y) & \implies f(t^6f(t)^2y)=f(t)^6 t^3 f(ty) \\ & \implies t^3 f(t \cdot yt^6)=t^3 f(t)^6 f(ty) \\ & \implies f(t^6(yt))=f(t)^6f(yt).\end{align*}So, $f(x^6y)=f(x)^6f(y)$ for all positive rationals $x,y$. This implies, in particular, that $f(x^6)=f(x)^6$. So, $$f(x^6y^6)=f(x)^6f(y^6) \implies f(xy)^6=f(x)^6f(y)^6 \implies f(xy)=f(x)f(y).$$This proves the claim. //// Note that we had $f(f(x)^2)=x^3f(x)$ which implies that $f(f(x))=x \sqrt{x f(x)}$. Let $p$ be a prime number. Then, $f(f(p))=p \sqrt{pf(p)}$ implies that $f(p)=r^2/p$ for some positive rational $r$. So $$f\left(\frac{r^2}p\right)=p\sqrt{p\cdot \frac{r^2}p} \implies \frac{f(r)^2}{f(p)}=pr \implies f(r)^2\cdot \frac{p}{r^2}=pr \implies f(r)^2=r^3.$$Note that $f(r)^2=r^3$ implies that $\sqrt{r}=r'$ is a positive rational. So $f(r')^4=(r')^6$ i.e. $f(r')^2=(r')^3$. This, again, implies that $r'$ is a perfect square of a rational i.e. $r$ is a perfect fourth power of a rational. Now it is evident, by induction, that for every $n=0,1,2,\dots$ there is a rational $m$ such that $r=m^{2^n}$. So $r$ must be $1$. It follows that $f(p)=\frac 1p$ for every prime $p$. Hence $f(x)=\frac 1x$ for every $x\in \mathbb{Q_+}$, since $f$ is multiplicative. This function clearly works.