I came across this super interesting question (Chinese Team Selection Test 1988, Problem 2) which I have not quite seen something like this and not quite sure where to start. Any suggestions/ help would be much appreciated. Thanks!
Find all functions $f: \mathbb{Q} \to \mathbb{C}$ satisfying
- For any $x_1, x_2, \ldots, x_{1988} \in \mathbb{Q}$, $f(x_1 + x_2 + \dots + x_{1988}) = f(x_1)f(x_2) \dots f(x_{1988})$.
- $\overline{f(1988)}f(x) = f(1988)\overline{f(x)}$ for all $x \in \mathbb{Q}$.
Write $n$ for $1988$.
One obvious solution is $f = 0$, the constant zero function. We exclude this case in the following.
Setting $x_2 = \cdots = x_n = 0$, we get $f(x_1) = f(x_1)f(0)^{n - 1}$. Choosing $x_1$ such that $f(x_1) \neq 0$, we get $f(0)^{n - 1} = 1$.
We write $a = f(0)$. It is an $(n - 1)$-st root of unity, hence nonzero.
Setting $x_3 = \cdots = x_n = 0$, we get $f(x_1 + x_2) = f(x_1)f(x_2)a^{n - 2} = f(x_1)f(x_2)a^{-1}$. In particular, setting $x_2 = -x_1$ shows that $f(x) \neq 0$ for any $x\in \Bbb Q$.
We define $g(x) = a^{-1}f(x)$. The above identity then translates to $g(x_1 + x_2) = g(x_1)g(x_2)$. Taking absolute values gives $|g(x_1 + x_2)| = |g(x_1)|\cdot|g(x_2)|$.
We write $b = |g(1)|$, a positive real number. It follows that $|g(k)| = b^k$ for any integer $k$, and then $|g(x)| = b^x$ for any rational number $x$.
From the identity $\overline{f(n)}f(x) = f(n)\overline{f(x)}$, we get $\overline{g(n)}g(x) = g(n)\overline{g(x)}$. Multiplying both sides by $g(n)g(x)$ gives $|g(n)|^2 g(x)^2 = g(n)^2|g(x)|^2$, or $b^{2n}g(2x) = b^{2x}g(2n)$.
We rewrite the last identity as $g(2x - 2n) = b^{2x - 2n}$. This being true for all $x \in \Bbb Q$, we have $g(x) = b^x$ for all $x\in \Bbb Q$.
Thus $f(x) = a\cdot b^x$, where $a$ is an $(n - 1)$-st root of unity and $b$ is a positive real number.
Conversely, it is easy to check that any function $f$ of this form satisfies the two conditions in the question. Therefore (together with the zero function) we have found all of them.