Problem 2031

We have $f(1997) = 1997$.

Label the equation as follows.
$$f(f(m) + f(n)) = m + n \quad (1)$$
If $f(a) = f(b)$ for some $a, b \in \mathbb{Z}^+$, then by putting $m = a$ and $m = b$ in (1), we obtain
$$a + n = f(f(a) + f(n)) = f(f(b) + f(n)) = b + n.$$
This implies $a = b$. So $f$ is injective. Now, using (1), we have
$$f(f(m + 1) + f(1)) = (m + 1) + 1 = m + 2 = f(f(m) + f(2)).$$
Using the injectivity, this becomes
$$f(m + 1) + f(1) = f(m) + f(2).$$
Let $d = f(2) - f(1)$. By induction, we can easily prove that
$$f(m) = f(1) + (m - 1)d$$
for any $m \in \mathbb{Z}^+$. In other words, $f(m) = am + b$ for some integers $a, b$. Now, we have
$$f(f(m) + f(n)) = a(am + b + an + b) + b = a^2m + a^2n + (2ab + b).$$
This is equal to $m+n$ for any $m, n \in \mathbb{Z}^+$ if and only if $a^2 = 1$ and $2ab + b = 0$. Clearly, the only solution with $a \ge 0$ is $(a, b) = (1, 0)$. This means $f(m) = m$ for any $m \in \mathbb{Z}^+$. In particular, we have $f(1997) = 1997$.