Let $f^r(x)$ denote the result when $f$ is applied to $f^{r-1}(x)$ , where $f^1(x)=f(x)$ . $\hfill \break \hfill \break$ If $f(p)=f(q)$ , then $f^2(p)=f^2(q)$ and $f^{f(p)}(p)=f^{f(q)}(q)$
$\implies p^2=f^2(p)\cdot f^{f(p)}(p)=f^2(q)\cdot f^{f(q)}(q)=q^2$
$\implies p=\pm q$
$\implies p=q$ since $p,q>0$ .
Therefore, $f$ is injective. It follows that $f^r$ is also injective.
Lemma 1: If $f^r(b)=a$ and $f(a)=a$ , then $b=a$ .
Proof:
$f^r(b)=a=f^r(a)$ which implies $b=a$ by injectivity of $f^r$ .
Lemma 2: If $f^2(m)=f^{f(m)}(m)=m$ , and $m$ is odd, then $f(m)=m$ .
Proof:
Let $f(m)=k$ . Since $f^2(m)=m$ , $f(k)=m$ . So, $f^2(k)=k$ . $\newline f^2(k)\cdot f^{f(k)}(k)=k^2$ .
Since $k\neq0$ , $f^{f(k)}(k)=k$
$\implies f^m(k)=k$
$\implies f^{gcd(m, 2)}(k)=k$
$\implies f(k)=k$
This proves Lemma 2.
I claim that $f(m)=m$ for all odd $m$ .
Otherwise, let $m$ be the least counterexample.
Since $f^2(m)\cdot f^{f(m)}(m)=m^2$ , either
$(1) f^2(m)=k<m$ , contradicted by Lemma 1 since $k$ is odd and $f^2(k)=k$ .
$(2) f^{f(m)}(m)=k<m$ , also contradicted by Lemma 1 by similar logic.
$(3) f^2(m)=m$ and $f^{f(m)}(m)=m$ , which implies that $f(m)=m$ by Lemma 2.
This proves the claim.
By injectivity, $f(1000)$ is not odd.
I will prove that $f(1000)$ can be any even number, $x$ . Let $f(1000)=x, f(x)=1000$ , and $f(k)=k$ for all other $k$ . If $n$ is equal to neither $1000$ nor $x$ , then $f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2$ . This satisfies the given property.
If $n$ is equal to $1000$ or $x$ , then $f^2(n)\cdot f^{f(n)}(n)=n\cdot n=n^2$ since $f(n)$ is even and $f^2(n)=n$ . This satisfies the given property.

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