orl wrote: Find all functions $f: \mathbb{Q} \mapsto \mathbb{C}$ satisfying (i) For any $x_1, x_2, \ldots, x_{1988} \in \mathbb{Q}$, $f(x_{1} + x_{2} + \ldots + x_{1988}) = f(x_1)f(x_2) \ldots f(x_{1988})$. (ii) $\overline{f(1988)}f(x) = f(1988)\overline{f(x)}$ for all $x \in \mathbb{Q}$. at the beginning, we suppose $f(x)\neq0$ for all $x \in \mathbb{Q}$. by $\overline{f(1988)}f(x) = f(1988)\overline{f(x)}$ we find $\frac{f(x)}{\overline{f(x)}}=\frac{f(1998)}{\overline{f(1998)}}\longrightarrow\frac{f(x)}{\overline{f(x)}}=\frac{f(y)}{\overline{f(y)}}=$costant for all x,y. This means that, if $f(x)=p(x)e^{i\theta(x)}$, $\frac{p(x)e^{i\theta(x)}}{p(x)e^{-i\theta(x)}}=e^{2i\theta(x)}=$costant, so $2i\theta(x)=$costant $(mod$ $2\pi)$, $\theta(x)=$costant $(mod$ $\pi)$: if we make $p(x)$ vary in the positive and negative reals, that means that $\theta(x)$ is costant. now consider the first relation: if we put $x_1=x_2=\ldots=x_{1998}=0$, we have $f(0)^{1998}=f(0)$, so f(0) is one of the 1997 complex 1997-roots of the unity, $f(0)=e^{\frac{2ki\pi}{1997}}$ (and this is $\theta(x)$ for all x) now let's put $x_1=x_2=\ldots=x_{1998}=a$, and then we have $f(a)^{1998}=f(1998a)$ again, let's put $x_3=x_4=\ldots=x_{1998}=a$, $x_1=a+b$, $x_2=a-b$ , from which we have $f(a)^{1996}f(a+b)f(a-b)=f(1998a)$. From this two relations we have $f(a)^2=f(a+b)f(a-b)$, changing $f(\frac{x+y}{2})^2=f(x)f(y)$,for all x,y. This means also that $f(\frac{x+y}{2})^2=f(x+y)f(0)$, and then $f(x+y)f(0)=f(x)f(y)$. With the substitution $f(x)=p(x)e^{i\theta(x)}=p(x)e^{\frac{2ki\pi}{1997}}$ we have $p(x)e^{\frac{2ki\pi}{1997}}p(y)e^{\frac{2ki\pi}{1997}}=p(x+y)e^{\frac{2ki\pi}{1997}}e^{\frac{2ki\pi}{1997}}$ $p(x)p(y)=p(x+y)$, with $p(x)$ function $p: \mathbb{Q} \mapsto \mathbb{R}$ and $p(0)=1$. This functional equation is trivial: the solutions are $p(x)=a^x$ for $a>0$, from which $f(x)=a^xe^{\frac{2ki\pi}{1997}}$ ($k\in\mathbb{Z}/_{1997}$). It can be easily seen that these functions satisfy the two text relations. Now suppose that there is an y such that $f(y)=0$. in the first relation, if we put $x_1=y$, $x_2=z-y$, $x_3=x_4=\dots=x_{1998}=0$, we see that $f(z)=f(y)f(z-y)f(0)^{1996}=0$ for all $z\in\mathbb{Q}$. so $f(x)=0$ is another solution. this ends the demonstration. hi everyone EDIT: exscuse me, i treated 1988 as 1998....