IMO 1998 · Problem 6

Consider all functions \(f\) from the set \(N\) of all positive integers into itself satisfying \(f(t^2f(s))=s(f(t))^2\) for all \(s\) and \(t\) in \(N\). Determine the least possible value of \(f(1998)\).

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{120} \]

Solution

Let \(c=f(1)\). Setting \(t=1\) in the given equation yields \[ f(f(n))=c^2n \qquad(n\in N). \tag{1} \] Consequently, \[ f(c^2n)=f(f(f(n)))=c^2f(n). \tag{2} \] Now substitute \(s=f(u)\) into the original equation. Using (1) and (2), we obtain \[ c^2f(t^2u)=f(c^2t^2u)=f(u)f(t)^2. \tag{3} \]

Define \(g(n)=f(n)/c\), initially a positive rational number. Then \(g(1)=1\), and (3) becomes \[ g(t^2u)=g(t)^2g(u). \tag{4} \] In particular, \(g(t^2)=g(t)^2\). Hence, for any positive integers \(a,b\), \[ g(ab)^2 =g(a^2b^2) =g(a)^2g(b^2) =g(a)^2g(b)^2. \] All values are positive, so \[ g(ab)=g(a)g(b). \tag{5} \]

We next show that every \(g(n)\) is an integer. Write \(g(n)=A/B\) in lowest terms. By (5), for every positive integer \(k\), \[ f(n^k)=c\,g(n)^k=\frac{cA^k}{B^k} \] is an integer. Since \(A^k\) and \(B^k\) are coprime, \(B^k\mid c\) for every \(k\). This forces \(B=1\).

Taking \(n=1\) in (1) gives \(f(c)=c^2\), so \(g(c)=c\). Therefore, using (1) and (5), \[ c^2n=f(f(n)) =c\,g(cg(n)) =c\,g(c)g(g(n)) =c^2g(g(n)). \] Thus \[ g(g(n))=n. \tag{6} \] In particular, \(g\) is injective, and \(g(n)>1\) whenever \(n>1\).

Moreover, \(g\) maps primes to primes. Indeed, if \(p\) were prime and \(g(p)=ab\) with \(a,b>1\), then \[ p=g(g(p))=g(a)g(b), \] where both factors exceed \(1\), a contradiction.

Since \[ 1998=2\cdot3^3\cdot37, \] the numbers \(p=g(2)\), \(q=g(3)\), and \(r=g(37)\) are three distinct primes. Consequently, \[ f(1998)=c\,pq^3r=c\,(pqr)q^2 \ge 1\cdot(2\cdot3\cdot5)\cdot2^2 =120. \]

To attain this bound, define \(F\) on primes by \[ F(2)=3,\quad F(3)=2,\quad F(5)=37,\quad F(37)=5, \] and \(F(p)=p\) for every other prime. Extend \(F\) completely multiplicatively to all positive integers, with \(F(1)=1\). Unique prime factorization makes this well-defined, and applying each prime swap twice shows that \(F(F(n))=n\). Hence \[ F(t^2F(s))=F(t)^2F(F(s))=sF(t)^2, \] so \(F\) satisfies the required equation. Finally, \[ F(1998)=F(2)F(3)^3F(37)=3\cdot2^3\cdot5=120. \] Therefore the least possible value is \(\boxed{120}\).

The solution was checked against Evan Chen's published IMO solution notes.