Let $N$ is the set of all positive integers. Determine all mappings $f: N-\{1\} \to N$ such that for every $n \ne m$ the following equation is true $$f(n)f(m)=f\left((nm)^{2021}\right)$$
Problem
Source: 2021 3nd Final Mathematical Cup Senior Division P1 FMC
Tags: functional, functional equation, algebra
30.10.2022 13:19
parmenides51 wrote: Let $N$ is the set of all positive integers. Determine all mappings $f: N-\{1\} \to N$ such that for every $n \ne m$ the following equation is true $$f(n)f(m)=f\left((nm)^{2021}\right)$$ For easier writing, let $a=2021$ Let $P(x,y)$ be the assertion $f(x)f(y)=f((xy)^a)$ true $\forall x,y\ge 2$, $x\ne y$ Comparaison of $P(x^{k+1},x)$ with $P(x^k,x^2)$ gives $f(x^{k+1})=f(x^k)\frac{f(x^2)}{f(x)}$ $\forall x\ge 2$, $\forall k\ge 3$ And so $f(x^k)=f(x^3)\left(\frac{f(x^2)}{f(x)}\right)^{k-3}$ $\forall x\ge 2$, $\forall k\ge 3$ So that $P(x,y)$ may also be written $Q(x,y)$ : $f(x)f(y)=f(x^3y^3)\left(\frac{f(x^2y^2)}{f(xy)}\right)^{a-3}$ $\forall x,y\ge 2$, $x\ne y$ Let $u,v\ge 3$, $u\ne v$ : $Q(x^u,x^v)$ is $f(x^u)f(x^v)=f(x^{3u+3v})\left(\frac{f(x^{2u+2v})}{f(x^{u+v})}\right)^{a-3}$ And so, using previously got expression for $f(x^k)$ : $f(x^3)\left(\frac{f(x^2)}{f(x)}\right)^{u-3}f(x^3)\left(\frac{f(x^2)}{f(x)}\right)^{v-3}$ $=f(x^3)\left(\frac{f(x^2)}{f(x)}\right)^{3u+3v-3}$ $\left(\frac{f(x^3)\left(\frac{f(x^2)}{f(x)}\right)^{2u+2v-3}}{f(x^3)\left(\frac{f(x^2)}{f(x)}\right)^{u+v-3}}\right)^{a-3}$ Which is $f(x^3)=\left(\frac{f(x^2)}{f(x)}\right)^{(u+v)(a-1)+3}$ And since this is true whatever are distinct $u,v\ge 3$, we get $f(x^2)=f(x)$ and $f(x^3)=1$ $\forall x>1$ And so $f(x^k)=1$ $\forall k\ge 3$ And $P(x,y)$ becomes $f(x)f(y)=1$ $\forall$ distinct $x,y\ge 2$ And so $\boxed{f(x)=1\quad\forall x\ge 2}$, which indeed fits.
30.10.2022 14:07
$c=2021$.For $x=yz^ct^{c^2}, y^cx^{c^2}z^{c^2}=x^cy^{c^2}z^{c^3}(t^{c-1})^{c^3} \implies f(y)f(x)f(z)=f(x)f(y)f(z)f(t^{c-1}) \implies f(t^{c-1})=1$ for all $t>1$. Because $c>3, f(m)f(m^{c-2})=f(m^{c(c-1)})=1$ gives $f(m)=1$ for all $m$. Nearly the same as APMO 2015 P2.
01.11.2022 03:41
Let $k=2021$. We have: $f((xyz)^{k^2})=f((xz)^k)f(y^k)=f(x)f(z)f(y^k) \forall x,y,z \in \mathbb{N}^* \setminus \lbrace 1 \rbrace$ $f((xyz)^{k^2})=f((yz)^k)f(x^k)=f(y)f(z)f(x^k) \forall x,y,z \in \mathbb{N}^* \setminus \lbrace 1 \rbrace$ $\Rightarrow \dfrac{f(x^k)}{f(x)}=\dfrac{f(y^k)}{f(y)} \forall x,y \in \mathbb{N}^* \setminus \lbrace 1 \rbrace$ Let $f(x^k)=cf(x) \forall x \in \mathbb{N}^* \setminus \lbrace 1 \rbrace$ We have: $c^2f(x)f(y)=f(x^k)f(y^k)=f((xy)^{k^2})=cf((xy)^k)=cf(x)f(y) \forall x,y \in \mathbb{N}^* \setminus \lbrace 1 \rbrace$ $\Rightarrow c=1$ $\Rightarrow f(x)f(y)=f((xy)^k)=f(xy)$ Therefore, $f(x)=f(x^k)=f(x \cdot x^{k-1})=f(x)f(x^{k-1})=f^2(x)f(x^{k-2})$ $\Rightarrow f(x)f(x^{k-2})=1 \forall x \in \mathbb{N}^* \setminus \lbrace 1 \rbrace$ $\Rightarrow f(x) \mid 1 \forall x \in \mathbb{N}^* \setminus \lbrace 1 \rbrace$ or $f(x)=1 \forall x \in \mathbb{N}^* \setminus \lbrace 1 \rbrace$