Iranian Mathematical Olympiad

First, by substituting $b = c = 0$ into the given relation, we obtain:
$$f(a) \cdot \bar{f}(a) \cdot \bar{f}(a) \cdot f(a) = 1 \Rightarrow |f(a)|^2 \cdot |f(a)|^2 = 1 \Rightarrow |f(a)|^4 = 1 \Rightarrow |f(a)| = 1$$
Since for every $x \in \mathbb{C}$, $x\bar{x} = |x|^2$, and we know $|f(x)| = 1$ implies $\bar{f}(x) = 1/f(x)$. The original relation is $f(a)\overline{f(a+b)f(a+c)f(a+b+c)} = 1$. Substituting $\bar{f}(x) = 1/f(x)$:
$$f(a) \frac{1}{f(a+b)} \frac{1}{f(a+c)} f(a+b+c) = 1$$
$$\Rightarrow f(a)f(a+b+c) = f(a+b)f(a+c)$$
Now, if $a = 0, c = 1$, we can write:
$$f(b+1) = \frac{f(b)f(1)}{f(0)} \Rightarrow f(n) = \frac{f(1)^n}{f(0)^{n-1}} = f(0) \left(\frac{f(1)}{f(0)}\right)^n$$
Since $f(n+T) = f(n)$, we have that $\frac{f(1)}{f(0)}$ is a root of unity. Therefore, every solution is of the form $f(n) = cw^n$ where $|c| = 1$ and $w$ is a root of unity. By substitution, we find that all such functions are solutions. ■