Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

No one?

Do the brackets indiciate floor() or just another layer of parentheses?

when $f: \mathbb{N}\rightarrow \mathbb{N}$ it's easy to see that it's parentheses

Denote $f(1)=a$, and put $m=n=1$, therefore $f(f(k))=a^{2}k$ and $f(ak^{2})=f^{2}(k)$, $\forall k \in \mathbb{N}$. Thus now, we have: $f^{2}(x) f^{2}(y)=f^{2}(x)f(ay^{2})=f(x^{2}f(f(ay^{2})))=$ $=f(x^{2}a^{3}y^{2})=f(a(axy)^{2})=f^{2}(axy)$ $\iff f(axy)=f(x)f(y) \Rightarrow f(ax)=af(x)$ $\iff af(xy)=f(x)f(y) , \forall x,y \in \mathbb{N}$. Now we can easily prove that $f(x)$ is divisible by $a$ for each $x$, more likely we have that $f^{k}(x)=a^{k-1}\cdot f(x^{k})$ is divisible by $a^{k-1}$. For proving the above asertion we consider $p^{\alpha}$ and $p^{\beta}$ the exact powers of a prime $p$ that tivide $f(x)$ and $a$ respectively, therefore $k\alpha \geq (k-1)\beta , \forall k \in \mathbb{N}$, therefore $\alpha\geq \beta$, so $f(x)$ is divisible by $a$. Now we just consider the function $g(x)=\frac{f(x)}{a}$. Thus: $g(1)=1, g(xy)=g(x)g(y), g(g(x))=x$. Since $g(x)$ respects the initial condition of the problem and $g(x)\leq f(x)$, we claim that it is enough to find the least value of $g(1998)$. Since $g(1998)=g(2 \cdot 3^{3}\cdot 37) =g(2) \cdot g^{3}(3)\cdot g(37)$, and $g(2),g(3),g(37)$ are disting prime numbers (the proof follows easily), we have that $g(1998)$, is not smaller than $2^{3}\cdot 3 \cdot 5=120$. But $g$ beeing a bijection, the value $120$, is obtained for any $g$, so we have that $g(2)=3, g(3)=2, g(5)=37, g(37)=5$, therefore the answer is $120$, and thus the problem is solved!

13 year bump?! We claim the minimum value of $f(1998)$ is $\boxed{120}$. Let $P(m, n)$ denote the given assertion. Additionally, let $f^k(x)$ denote $f$ applied $k$ times on $x$, let $f(x)^k$ denote $f(x)$ to the $k$'th power, and let $c = f(1)$. Suppose $f$ is such that $f(1998)$ is minimal. From $P(m, 1)$, we get \begin{align*} f^2(m) &= mc^2. \end{align*}From $P(1, m)$, we get \begin{align*} f(cm^2) &= f(m)^2. \end{align*}From $P(f(cm^2), n)$, we get \begin{align*} f(n^2f(f(cm^2))) &= f(cm^2)f(n)^2 \\ f(c^3m^2n^2) &= f(m)^2f(n)^2 \\ f(cmn)^2 &= f(m)^2f(n)^2 \\ f(cmn) &= f(m)f(n). \end{align*}Setting $n = 1$ in the above, we obtain \begin{align*} f(cm) &= cf(m). \end{align*}Hence, for all $m, n$ we have \begin{align*} cf(mn) &= f(m)f(n). \end{align*}In particular, setting $m = n$ here yields $c \mid f(m)^2$ for all $m$. Suppose $c > 1$, and let $p$ be a prime dividing $c$. Note that, for all $m$, we must have $\nu_p(f(m)) \geq \frac{\nu_p(c)}{2}$. Suppose there exists $m$ for which $\nu_p(f(m)) < \nu_p(c)$. Then, we have \begin{align*} cf(m^2) &= f(m)^2 \\ \nu_p(c) + \nu_p(f(m^2)) &= 2\nu_p(f(m)) \\ \nu_p(f(m^2)) &= 2\nu_p(f(m)) - \nu_p(c) \\ &< \nu_p(f(m)). \end{align*}Iterating this, we find that for large enough $k$ we have $\nu_p(f(m^{2^k})) < \frac{\nu_p(f(m))}{2}$, a contradiction. Therefore, we have $\nu_p(f(m)) = \nu_p(c)$ for all primes $p \mid c$, implying that $c \mid f(m)$ for all $m$. Now, let $g(m) = \frac{f(m)}{c}$. We rewrite $P(m, n)$ as follows: \begin{align*} cf\left(n^2\cdot \frac{f(m)}{c}\right) &= mc^2\left(\frac{f(n)}{c}\right)^2 \\ c^2g(n^2g(m)) &= mc^2g(n)^2 \\ g(n^2g(m)) &= mg(n)^2, \end{align*}so $g$ satisfies $P$ as well. In the case that $c > 1$, we have that $g(1998) < f(1998)$, contradicting the fact that $f$ minimizes $f(1998)$. Therefore, we have $c = 1$, so that \begin{align*} f(mn) &= f(m)f(n) \end{align*}for all $m, n$, and $f^2(m) = m$. We are now nearly done. Let $p$ be a prime, and let $f(p) = \prod_{i = 1}^k p_i^{e_i}$ for distinct primes $p_i$ and positive integers $e_i$. We have \begin{align*} f^2(p) &= f\left(\prod_{i = 1}^k p_i^{e_i}\right) \\ p &= \prod_{i = 1}^k f(p_i)^{e_i}. \end{align*}If $k > 1$, it follows that there exists a prime $q$ for which $f(q) = 1$. In this case, we have $f^2(q) = f(1) = 1$, contradicting the fact that $f^2(q) = q$. Hence, $k = 1$, and we have $p = f(q^r) = f(q)^r$ for some prime $q$ and positive integer $r$. Clearly, this implies that $r = 1$, so $f(p) = q$. Thus, $f$ is a multiplicative involution which maps primes to primes. We claim that all such functions $f$ satisfy the given functional equation. Indeed, for any such $f$, we have \begin{align*} f(n^2f(m)) &= f(n^2)f^2(m) \\ f(n^2f(m)) &= mf(n)^2. \end{align*} To finish, we have \begin{align*} f(1998) &= f(2 \cdot 3^3 \cdot 37) \\ &= f(3)^3f(2)f(37) \\ &\leq 2^3 \cdot 3 \cdot 5 \\ &= 120, \end{align*}as claimed. $\Box$

Let $P(m,n)$ denote the FE. The key identities we'll use are $P(m,1)$ gives $f(f(m))=mf(1)^2$. $P(1,n)$ gives $f(f(1)n^2)=f(n)^2$. Consider $S=f\big(a^2f(f(1)b^2)\big)$. Note $f(f(1)b^2)=f(b)^2$, so $S=f(a^2f(b)^2)$. On the other hand, $S=f(1)b^2f(a)^2$ by using the FE. Hence, \[ f(a^2f(b)^2)=f(1)b^2f(a)^2. \]Taking $f$ of both sides gives and using the two bulleted identities gives \[ a^2f(b)^2 f(1)^2 = f(bf(a))^2 \implies af(b)f(1)=f(bf(a)). \]Substitute $f(b)$ for $b$ above to get $af(f(b))f(1)=f(f(b)f(a))$, i.e. $abf(1)^3=f(f(b)f(a))$. Taking $f$ of both sides gives \[ f(abf(1)^3) = f(b)f(a)f(1)^2.\]Plug in $b=1$ above to get $f(af(1)^3)=f(a)f(1)^3$. So $f(abf(1)^3)=f(ab)f(1)^3$, which means \[ f(ab)f(1)^3 = f(b)f(a)f(1)^2 \implies \boxed{f(1)f(ab)=f(a)f(b)}. \] Now we claim $f(1)\mid f(x)$ for all $x$. We have $f(a^k)=f(a)^k/f(1)^{k-1}$ by iterating the above. Consider some $p\mid f(1)$, then $k\nu_p(f(a)) \ge (k-1)\nu_p(f(1))$, i.e. $\nu_p(f(a)) \ge \tfrac{k-1}{k}\nu_p(f(1))$. Since this holds for all $k\ge 1$, actually $\nu_p(f(a)) \ge \nu_p(f(1))$. Therefore $f(1)\mid f(a)$, as claimed. Now let $g(x)=f(x)/f(1)$ be an integer function. Then $g(ab)=g(a)g(b)$. We claim $g(g(m))=m$. Indeed, \[ mf(1)^2=f(f(m))=f(1)g(f(m))=f(1)g(f(1)g(m))=f(1)g(f(1))g(g(m)). \]But $g(f(1))=f(f(1))/f(1)$, and since $P(1,1)$ gives $f(f(1))=f(1)^2$, actually $g(f(1))=f(1)$. Hence, the above equalities give $mf(1)^2=f(1)^2g(g(m))$, i.e. $g(g(m))=m$. Now, we simply pair up primes $(p,q)$ and set $f(p)=q$, $f(q)=p$, and then since $g$ is multiplicative, all other outputs are determined. It is easy to check that $g$ is involutive in general in this construction (each prime in the prime factorization involutes). This describes the general function $g$. Finally, to minimize $f(1998)$, set $f(1)=1$, and then \begin{align*} f(1998) &= g(1998) = g(2\cdot 3^3\cdot 37) = g(2)g(3)^3g(37) \ge 3\cdot 2^3 \cdot 5 = 120, \end{align*}where we set $g(2)=3,g(3)=2$ and $g(37)=5$.

We first provide a construction for all functions $f$, then prove all functions $f$ are of this form. Let $d$ be an arbitrary positive integer and $g$ be an involutive multiplicative function that takes primes to primes and fixes $d$. Then, define $f(n)=dg(n)$. It is not too hard to check $f$ works. Let $P(m,n)$ denote the condition for $m,n$. Let $d=f(1)$. Observe that By $P(1,1)$, $f(d)=d^2$. By $P(n,1)$, $f(dn^2)=f(n)^2$. By $P(1,n)$, $f(f(n))=d^2n$. Now, note that \[f(a)^2f(b)^2=f(b^2f(f(a)^2))=f(b^2f(f(da^2)))=f(d^3a^2b^2)=f(dab)^2,\]hence \[f(a)f(b)=f(dab)\qquad (\diamondsuit).\]Take $a=1$ in $\diamondsuit$ to yield $df(n)=f(dn)$, so $\diamondsuit$ rewrites as \[f(a)f(b)=df(ab)\qquad (\heartsuit).\]Note that the information we have extracted implies the given equation because \[f(n^2f(m))=\frac{1}{d}f(n^2)f(f(m))=mf(n)^2,\]so we ignore it now. Claim: Every $f(n)$ is divisible by $d$. Solution: Note \[f(n^k)=\frac{1}{d}f(n)f(n^{k-1})=\cdots = \frac{1}{d^{k-1}}f(n)^k,\]which implies $d\mid f(n)$ by taking large $k$ and analyzing the $p$-adic valuation of $f(n),d$. $\fbox{}$ Now, let $g(n)=f(n)/d$ be a function from $\mathbb{N}\to\mathbb{N}$, which exists by the previous claim. Our information rewrites as $g(d)=d$, $g(a)g(b)=g(ab)$, $g(g(n))=n$. By multiplicativity, $g$ preserves the number of divisors of a number, hence $g$ takes primes to primes, and we are done by taking $g$ and $d$ in our construction to be as extracted. Now, we actually solve the problem; prime factorize $1998$ as $2\cdot3^3\cdot37$, so $g(1998)\ge 2^3\cdot3\cdot5=120$.

pohoatza wrote: Denote $f(1)=a$, and put $m=n=1$, therefore $f(f(k))=a^{2}k$ and $f(ak^{2})=f^{2}(k)$, $\forall k \in \mathbb{N}$. Thus now, we have: $f^{2}(x) f^{2}(y)=f^{2}(x)f(ay^{2})=f(x^{2}f(f(ay^{2})))=$ $=f(x^{2}a^{3}y^{2})=f(a(axy)^{2})=f^{2}(axy)$ $\iff f(axy)=f(x)f(y) \Rightarrow f(ax)=af(x)$ $\iff af(xy)=f(x)f(y) , \forall x,y \in \mathbb{N}$. Now we can easily prove that $f(x)$ is divisible by $a$ for each $x$, more likely we have that $f^{k}(x)=a^{k-1}\cdot f(x^{k})$ is divisible by $a^{k-1}$. For proving the above asertion we consider $p^{\alpha}$ and $p^{\beta}$ the exact powers of a prime $p$ that tivide $f(x)$ and $a$ respectively, therefore $k\alpha \geq (k-1)\beta , \forall k \in \mathbb{N}$, therefore $\alpha\geq \beta$, so $f(x)$ is divisible by $a$. Now we just consider the function $g(x)=\frac{f(x)}{a}$. Thus: $g(1)=1, g(xy)=g(x)g(y), g(g(x))=x$. Since $g(x)$ respects the initial condition of the problem and $g(x)\leq f(x)$, we claim that it is enough to find the least value of $g(1998)$. Since $g(1998)=g(2 \cdot 3^{3}\cdot 37) =g(2) \cdot g^{3}(3)\cdot g(37)$, and $g(2),g(3),g(37)$ are disting prime numbers (the proof follows easily), we have that $g(1998)$, is not smaller than $2^{3}\cdot 3 \cdot 5=120$. But $g$ beeing a bijection, the value $120$, is obtained for any $g$, so we have that $g(2)=3, g(3)=2, g(5)=37, g(37)=5$, therefore the answer is $120$, and thus the problem is solved! Can anyone tell HOW $\alpha ≥ \beta $?

abvk1718 wrote: pohoatza wrote: Denote $f(1)=a$, and put $m=n=1$, therefore $f(f(k))=a^{2}k$ and $f(ak^{2})=f^{2}(k)$, $\forall k \in \mathbb{N}$. Thus now, we have: $f^{2}(x) f^{2}(y)=f^{2}(x)f(ay^{2})=f(x^{2}f(f(ay^{2})))=$ $=f(x^{2}a^{3}y^{2})=f(a(axy)^{2})=f^{2}(axy)$ $\iff f(axy)=f(x)f(y) \Rightarrow f(ax)=af(x)$ $\iff af(xy)=f(x)f(y) , \forall x,y \in \mathbb{N}$. Now we can easily prove that $f(x)$ is divisible by $a$ for each $x$, more likely we have that $f^{k}(x)=a^{k-1}\cdot f(x^{k})$ is divisible by $a^{k-1}$. For proving the above asertion we consider $p^{\alpha}$ and $p^{\beta}$ the exact powers of a prime $p$ that tivide $f(x)$ and $a$ respectively, therefore $k\alpha \geq (k-1)\beta , \forall k \in \mathbb{N}$, therefore $\alpha\geq \beta$, so $f(x)$ is divisible by $a$. Now we just consider the function $g(x)=\frac{f(x)}{a}$. Thus: $g(1)=1, g(xy)=g(x)g(y), g(g(x))=x$. Since $g(x)$ respects the initial condition of the problem and $g(x)\leq f(x)$, we claim that it is enough to find the least value of $g(1998)$. Since $g(1998)=g(2 \cdot 3^{3}\cdot 37) =g(2) \cdot g^{3}(3)\cdot g(37)$, and $g(2),g(3),g(37)$ are disting prime numbers (the proof follows easily), we have that $g(1998)$, is not smaller than $2^{3}\cdot 3 \cdot 5=120$. But $g$ beeing a bijection, the value $120$, is obtained for any $g$, so we have that $g(2)=3, g(3)=2, g(5)=37, g(37)=5$, therefore the answer is $120$, and thus the problem is solved! Can anyone tell HOW $\alpha ≥ \beta $? We know that $\frac{\alpha}{\beta} \geq \frac{k-1}{k}$. Let $X=\left\{ \frac{k-1}{k} | k \in \mathbb{N} \right\}$. Then $\sup X=1$. But $\frac{\alpha}{\beta}$ is a upper bound of $X$. Therefore, $\frac{\alpha}{\beta} \geq 1$.

Good problem.

Let $c=f(1).$ Then, $$m=n=1\implies f(f(m))=c^2m \text{ and } f(cn^2)=f^2(n).$$It follows that \begin{align*} f^2(m)f^2(n)=f^2(m)f(cn^2)\\=f(m^2f(f(cn)))=f(m^2c^3n^2)\\=f^2(cmn) \implies f(m)f(n)=f(cmn).\end{align*}Also $$f(cm)=cf(m)\implies cf(mn)=f(m)f(n).~~~~~(*)$$ Claim 1: $c|f(n)~~ \forall n\in \mathbb{N}.$ Proof. Let $p$ be any prime and let $s\geq 0$ and $r\geq 0$ be the degrees of $p$ in the prime representation of $c$ and $f(n),$ respectively. By induction from $(*)$, $$f^t(n)=c^{t-1}f(n^t) ~~\forall t \in \mathbb{N}.$$It follows that $r\geq s,$ which provides the claim. $\blacksquare$ Let $g(n)=\frac{f(n)}{c},$ then \begin{align*} g(mn)=g(m)g(n) ~~~~~(\alpha)\\ g(g(m))=m ~~~~~(\beta) \\ \forall m,n \in \mathbb{N}.\end{align*}Setting $m=n=1$ in $(\alpha) \implies g(1)=1 ~~~~~(\gamma).$ Claim 2: $g(p)$ is a prime whenever $p$ is a prime. Proof. Let $g(p)=ab.$ From $(\alpha)$ and $(\beta)$, $$p=g(g(p))=g(ab)=g(a)g(b).$$So we can assume, $g(u)=1 \implies u=1.$ As desired. $\blacksquare$ Now let $n\geq 2$ and $n=p_1^{t_1}\dots p_k^{t_k}$ be the prime representation of $n.$ Then from $(\alpha)$, $$g(n)=g^{t_1}(p_1)\dots g^{t_k}(p_k).~~~~~(**)$$It follows that the function $f$ is determined solely by $c=f(1)$ and the function $g$ mapping from the set of primes to the set of primes s.t. $g(g(p))=p.$ Claim 3: $f(1998)=120.$ Proof. Notice that $$f(1998)=f(2\cdot 3^3\cdot 37)=f(1)g(2)g^3(3)g(37).$$Because $g(2),g(3),g(37)$ are distinct primes we have $$g(2)g^3(3)g(37)\geq 3\cdot 2^3 \cdot 5=120\implies f(1998)\geq 120.$$It is easy to check that the inequality is achievable when $f(1)=1, g(2)=3,g(3)=2,g(37)=5$ and $g(p)=p~~\forall p \neq 2,3,37.$ Thus the proof is complete. $\blacksquare$

The answer is $f(n) = cg(n)$, such that $g\colon \mathbb{N}\to \mathbb{N}$ maps primes to primes, $g$ is an involution, $g$ is multiplicative, and $g(c) = c$. These can be checked to work. Let $P(m,n)$ denote the given assertion and $f(1) = c$. $P(m,1): f(f(m)) = m\cdot c^2$. From here we see that $f$ is injective. $P(1,m): f(m^2 \cdot c) = f(m)^2$. $P(1,1): f(c) = c^2$. $P(m,c): f(f(m)c^2) = m c^4$ $P(c,m): f(m^2c^2) = c f(m)^2$. $P(f(m),n): f(m c^2 n^2 ) = f(m) f(n)^2$. $P(m^2c, f(n)): f(f(m)^2 f(n)^2) = m^2 n^2 c^5$. $P(f(m),c): f(mc^4) = c^4f(m)$. So \[f(m^2n^2c^5) = f(c^4 ((mn)^2 c)) = c^4f(mn)^2 \] Thus $f(m)^2 f(n)^2 c^2 = c^4 f(mn)^2$, which implies \[f(m)f(n) = cf(mn)\] By induction we find \[f(m^k) = \frac{f(m)^k}{c^{k-1}}\]for any $k$. Suppose we had a prime $p$ for which $\nu_p(c) > \nu_p(f(m))$. If $\nu_p(c) = a, \nu_p(f(m)) = b$, we have \[bk\ge a(k-1) = ak-a,\]so $(a-b)k \le a $, which implies $k \le \frac{a}{a-b}$. However, we can set $k$ arbitrarily large, contradiction. Thus $c\mid f(m)$ for all $m$. Let $g(m) = \frac{f(m)}{c}$. The equations \[f(m)f(n) = cf(mn) \text{ and } f(f(m)) = c^2 m\]imply that $g$ is multiplicative and involutive. Since $f(c) = c^2$, we have $g(c) = c$. It remains to show $g$ maps primes to primes. Note that if $g(m) = 1$ then $m=1$. So for any $x,y>1$, we have $g(xy) = g(x)g(y)$, which implies $g(n) $ is composite for all $n$. Since $g(g(p)) = p$, $g(p)$ can't be composite, so $g(p)$ is prime.

Denote by $P(m,n)$ the assertion. Let $f(1)=c$. Then, we have that \begin{enumerate}[(i)] \item $P(1,1)$ giving $f(c)=c^2$ \item $P(m,1)$ giving $f(f(m))=mc^2=mf(c)$. \item $P(1,n)$ giving $f(n^2c)=f(n)^2$ \end{enumerate} Note that $P(f(m),n)$ gives $f(m)f(n)^2=f(n^2f(f(m)))=f(n^2mc^2)$. Substituting $m\to m^2c$, we get \[f(m)^2f(n)^2=f(m^2c)f(n)^2=f(m^2n^2c^3)=f((mnc)^2c)=f(mnc)^2\]Thus, we have $f(m)f(n)=f(mnc)$. Letting $m=1$ in that, we have $f(cn)=cf(n)$, so $f(m)f(n)=cf(mn)$. Let $f(n)=cg(n)$ then $g(m)g(n)=g(mn)$, $g(1)=1$, and a manipulation of $P(m,1)$ gives $c^2g(g(n))=cg(cg(n))=c^2n$ so $g(g(n))=n$. Additionally, $g$ is bijective. $~$ We claim that $g(p)$ is prime for primes $p$. If not, then let $g(p)=xy$ where $x\ge y>1$, then \[p=g(g(p)=g(xy)=g(x)g(y)\]so $g(x)$ or $g(y)$ is one, a contradiction because $g(x)=1\iff x=1$. $~$ We have $f(1998)=cg(2)g(3)^3g(37)\ge 1\cdot 3\cdot 2^3\cdot 5 = 120$, achievable when $g$ just involutes the primes so that $2\to 3, 37\to 5$, and $c=1$.

Denote the assertion with $P(m, n)$. Let $d = f(1)$ and let $c = n^2f\left(\frac{1}{f(n)^2}\right)$ such that $f(c) = 1$. Then, $P(1, 1)$ gives that $f(d) = d^2$. Likewise, $P(n, 1)$ gives that $f(f(n)) = d^2n$ so $f$ is injective. Finally $P(1, n)$ means that $f(dn^2) = f(n)^2$ As such, \[ f(a)^2f(b)^2 = f(da^2)f(b)^2 = f(b^2f(f(da^2))) = f(b^2a^2d^3) = f(dab)^2 \]so $f(a)f(b) = f(dab)$ and thus $df(a) = f(da)$. Thus, this becomes $f(a)f(b) = df(ab)$. Thus, $d = \frac{f(a)f(b)}{f(ab)}$ so \[ \nu_p(f(a)) + \nu_p(f(b)) - \nu_p(f(ab)) = 1 \]If there is some $f(a)$ such $d \nmid f(a)$, then \[ \nu_p(f(b)) - \nu_p(f(ab)) = 1 \]However, then the sequence $f(b), f(ab), f(a^2b), \dots$ has ever decreasing $\nu_p$, contradiction. Thus, $d \mid f(n)$ for all $n$. Define $g(n) = \frac{f(n)}{d}$. Then we have that $d^2g(a)g(b) = d^2g(a)g(b)$ so $g$ is multiplicative. Furthermore, $g(dg(n)) = dn$ implies that $g(d)g(g(n)) = dn$ so $g(g(n)) = n$. We now characterize all multiplicative involutions. Note that $g(1) = 1$. Let $p_1, p_2, \dots$ be the primes and have $g(p_i) = q_i$, and note that $q_i \ne 1$ by injectivity. Then, $g(q_i) = p_i$ can only hold if $q_i$ is also a prime. Thus, for any such pair $(g, d)$, $f(n) = d \cdot g(n)$ satisfies the original equation. Thus $f(1998)$ is minimized when $d = 1$ and since $1998 = 2 \cdot 3^3 \cdot 37$, the minimal value is $2^3 \cdot 3 \cdot 5 = 120$.

We will first characterize all $f$. Let $C\in\mathbb{N}$ and let $P$ be the set of primes. Then let $h$ be an involution on $P$, and extend $h$ to be a strongly multiplicative function over $\mathbb{N}$. Then $f(n)=Ch(n)$ is the set of solutions. This obviously works. In particular, the set of $f(1998)=f(2\cdot 3^3\cdot 37)$ is $Cp^3 qr$ for a positive integer $C$ and pairwise distinct primes $p,q,r$. Thus the minimum possible value of $f(1998)$ is $2^3\cdot 3\cdot 5=\boxed{120}$. Let $f(1)=C$. $P(m,1)$ gives $f(f(m))=mC^2$. Thus $f(mC^2)=f(f(f(m)))=f(m)C^2$. $P(m,Cn)$ gives $f(C^2n^2f(m))=mf(Cn)^2$. Thus $C^2mf(n)^2=C^2f(n^2f(m))=mf(Cn)^2$, so $Cf(n)=f(Cn)$. $P(1,n)$ gives $f(Cn^2)=f(n)^2$, so $Cf(n^2)=f(n)^2$. Claim. For all $n\in\mathbb{N}$, $C\mid f(n)$. Proof. Let $f(n)=A$. Then $f(n^2)=\frac{A^2}{C}$, so $f(n^4)=\frac{A^4}{C^3}$, so $f(n^8)=\frac{A^8}{C^7}$, etc, so for all $m\in\mathbb{N}$, \[\frac{A^{2^m}}{C^{2^m-1}}\in\mathbb{N},\]so by $\nu_p$ reasons, $C\mid A$. Let $g(m):=\frac1C f(m)$. Then \[Cg(n^2Cg(m))=mC^2g(n)^2\Longrightarrow g(n^2g(m))=mg(n)^2.\]So we're just solving the original FE with the added condition $f(1)=1$. Our previous results give $f(f(n))=n$ and $f(n^2)=f(n)^2$. $P(f(m),n)$ gives $f(mn^2)=f(m)f(n)^2$. These three conditions are equivalent to the original problem(besides the $f(1)=1$ thing). Note that $f(m^3)=f(m)f(m)^2=f(m)^3$, $f(m^4)=f((m^2)m^2)=f(m^2)f(m)^2=f(m)^4$, etc, so $f(m^n)=f(m)^n$ for all $m,n$. Now consider two relatively prime numbers $p,q>1$. Let $f(p)=P$ and $f(q)=Q$. We would like to determine the value of $f(p^m q^n)$ for all $m,n\in\mathbb{Z}_{\ge 0}$. Firstly, $f(p^m)=P^m$ and $f(q^n)=Q^n$. Thus we have \[f(p^{2m} q^{2n})=f(p^{2m} (q^n)^2)=f(p^{2m})f(q^n)^2=P^{2m} Q^{2n}.\]Therefore, \[f(p^m q^n)^2=f(p^{2m} q^{2n})=P^{2m} Q^{2n}\Longrightarrow f(p^m q^n)=P^m Q^n.\]Thus we have proven that $f$ is strongly multiplicative. Thus the original problem is equivalent to "strongly multiplicative and involution"(besides the $f(1)=1$ thing). Note that $f$ of a composite number is composite. Therefore, a prime cannot map to a composite number or $1$(since $f$ is an involution), so primes map to primes. Thus we can pair up primes in a cycle of two or have self-loops, then the strongly multiplicativeness does the rest. $\blacksquare$