http://www.kalva.demon.co.uk/imo/isoln/isoln986.html Let f(1) = k. Then f$ (kt^2) = f(t)^2$ and $ f(f(t)) = k^2t$. Also $ f(kt)^2 = 1.f(kt)^2 = f(k^3t^2) = f(1^2f(f(kt^2))) = k^2f(kt^2) = k^2f(t)^2$. Hence $ f(kt) = k f(t)$. By an easy induction $ k^nf(t^{n + 1}) = f(t)^{n + 1}$. But this implies that k divides f(t). For suppose the highest power of a prime p dividing k is a > b, the highest power of p dividing f(t). Then a > b(1 + 1/n) for some integer n. But then na > (n + 1)b, so $ k^$n does not divide $ f(t)^{n + 1}$. Contradiction. Let $ g(t) = f(t)/k$. Then $ f(t^2f(s)) = f(t^2kg(s)) = k f(t^2g(s) = k^2g(t^2g(s))$, whilst s $ f(t)^2 = k^2s f(t)^2$. So $ g(t^2g(s)) = s g(t)^2$. Hence g is also a function satisfying the conditions which evidently has smaller values than f (for k > 1). It also satisfies g(1) = 1. Since we want the smallest possible value of f(1998) we may restrict attention to functions f satisfying f(1) = 1. Thus we have f(f(t) = t and f(t^2) = f(t)^2. Hence f(st)^2 = f(s^2t^2) = f(s^2f(f(t^2))) = f(s)2f(t^2) = f(s)^2f(t)^2. So f(st) = f(s) f(t). Suppose p is a prime and f(p) = m·n. Then f(m)f(n) = f(mn) = f(f(p)) = p, so one of f(m), f(n) = 1. But if f(m) = 1, then m = f(f(m)) = f(1) = 1. So f(p) is prime. If f(p) = q, then f(q) = p. Now we may define f arbitarily on the primes subject only to the conditions that each f(prime) is prime and that if f(p) = q, then f(q) = p. For suppose that $ s = p_1^{a_1}...p_r^{a_r}$ and that $ f(p_i) = q_i$. If t has any additional prime factors not included in the $ q_i$, then we may add additional p's to the expression for s so that they are included (taking the additional a's to be zero). So suppose $ t = q_1^{b_1}...q_r^{b_r}$.Then $ t^2f(s) = q_1^{2b_1 + a_1} ...q_r^{2b_r + a_r}$ and hence $ f(t^2f(s) = p_1^{2b_1 + a_1} ...p_r^{2b_r + a_r} = s f(t)^2$. We want the minimum possible value of f(1998). Now $ 1998 = 2.3^3.37$, so we achieve the minimum value by taking f(2) = 3, f(3) = 2, f(37) = 5 (and f(37) = 5). This gives $ f(1998) = 3.2^35 = 120$

Isn't this IMO 1998 Q6??

Solution