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Source: Iran 2024 3rd Round Test 1 P1

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M11100111001Y1R

#1 h Aug 23, 2024, 10:27 AM • 1 Y

Y by sami1618

Suppose that $T\in \mathbb N$ is given. Find all functions $f:\mathbb Z \to \mathbb C$ such that, for all $m\in \mathbb Z$ we have $f(m+T)=f(m)$ and:
$\forall a,b,c \in \mathbb Z: f(a)\overline{f(a+b)f(a+c)}f(a+b+c)=1.$ Where $\overline{a}$ is the complex conjugate of $a$ .

This post has been edited 3 times. Last edited by M11100111001Y1R, Aug 29, 2024, 11:03 AM

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pco

#2 h Aug 23, 2024, 12:02 PM • 1 Y

Y by sami1618

M11100111001Y1R wrote:

Suppose that $T\in \mathbb N$ is given. Find all functions $f:\mathbb Z \to \mathbb C$ such that, for all $m\in \mathbb Z$ we have $f(m+T)=f(m)$ and:
$\forall a,b,c \in \mathbb Z: f(a)\overline{f(a+b)f(a+c)}f(a+b+c)=1.$ Where $\overline{a}$ is the complex conjugate of $a$ .

Let $P(x,y,z)$ be thge assertion $f(x)\overline{f(x+y)}\overline{f(x+z)}f(x+y+z)=1$

$P(x,T,T)$ $\implies$ $f(x)^2\overline{f(x)}^2=1$ and so $|f(x)|=1$ $\forall x$
So $\overline{f(x)}=\frac 1{f(x)}$

And $P(x,y,z)$ becomes $f(x)f(x+y+z)=f(x+y)f(x+z)$
Switching there $x,y$ and dividing ; $\frac{f(x)}{f(y)}=\frac{f(x+z)}{f(y+z)}$
And so $\frac{f(x+z)}{f(x)}=\frac{f(y+z)}{f(y)}$ depends only on $z$ is some function $g(z)$

So $f(x+z)=f(x)g(z)$
Switching there $z,x$ and dividing, we get $\frac{g(x)}{f(x)}=\frac{g(z)}{f(z)}=c$ constant

So $g(x)=cf(x)$ and $f(x+z)=cf(x)f(z)$ and so $f(x+1)=cf(1)f(x)$ and so $f(x)=\alpha \beta^x$

Back to $|f(x)|=1$ , this gives $f(x)=e^{i(ax+b)}$
And $f(x+T)=f(x)$ implies $aT=2k\pi$

And so $\boxed{f(x)=e^{i\left(\frac{2k\pi}Tx+b\right)}\quad\forall x\in\mathbb Z_{>0}}$ , which indeed fits, whatever are $k\in\mathbb Z$ , $b\in\mathbb R$

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#3 h Aug 23, 2024, 1:00 PM • 2 Y

Y by pco, sami1618

After $f(x)f(x+y+z)=f(x+y)f(x+z)$ we can also get directly to $f(x)=\alpha\beta^x$ by just putting $y=z=1$ and induction.

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#4 h Aug 23, 2024, 7:58 PM • 1 Y

Y by qwerty123456asdfgzxcvb

Convert to Polar Coordinates, specifically $f(x)=(r(x),\theta(x))=re^{i \theta }$ with $r\in [0,\infty)$ and $\theta$ reckoned modulo $2\pi$ .

All functions must satisfy $r(x)=1$ for all $x$ and $\theta(x)\equiv \frac{2\pi c_1 x}{T}+c_2$ for some constant integer $c_1$ and some constant real number $c_2$ . These functions can be verified to work.

Re-writing the assertion in polar form, we must have $r(a)r(a+b)r(a+c)r(a+b+c)=1$ $\theta(a)+\theta(a+b+c)\equiv \theta(a+c)+\theta(a+b)$ Notice in the first we can plug in $b=c=0$ so we must have that $r(a)^4=1$ so $r(a)=1$ for all integers $a$ . For the second equation shifting $\theta(x)$ still works so assume that $\theta(0)=0$ and then letting $a=0$ gives a variant of Cauchy's Functional Equation which is well known to be satisfied for $\theta(x)\equiv cx$ . Thus $\theta(x)$ must be linear. At the same time $\theta(x)\equiv \theta(x+T) \pmod{2\pi}$ so we must have the leading coefficient of $\theta(x)$ to be a multiple of $2\pi /T$ leading to the above form.

This post has been edited 2 times. Last edited by sami1618, Aug 23, 2024, 8:13 PM

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