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 $k\in\mathbb Z$, $b\in\mathbb R$

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.

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.