Solution: We provide yet another (more technical) solution assuming Step 1 and Step 2 of Solution 1. By Claim 5 in Solution 1, every $a \geqslant 2$ is a descendant of 1 . Let $g$ and $h$ be the functions on $\mathbb{Z}_{\geqslant 2}$ such that $f^{g(a)}(1)=a$ and $h(a)=f(a-1)$. Then, $g: \mathbb{Z}_{\geqslant 2} \rightarrow \mathbb{Z}_{\geqslant 1}$ and $h: \mathbb{Z}_{\geqslant 2} \rightarrow \mathbb{Z}_{\geqslant 2}$ are bijections. The equation $P(a, b)$ can be rewritten as  $$ g(a h(b))=g(a)+(b-1) h(a) $$  Consider the set $S_{a}=g\left(a \cdot \mathbb{Z}_{>0}\right)$. Since $h$ is a bijection onto $\mathbb{Z}_{\geqslant 2}$, we have  $$ S_{a}=g(a)+h(a) \cdot \mathbb{Z}_{\geq 0} $$  Consider the intersection $S_{a} \cap S_{b}=S_{\operatorname{lcm}(a, b)}$. If we put $c=\operatorname{lcm}(a, b)$, this gives  $$ \left(g(a)+h(a) \cdot \mathbb{Z}_{\geqslant 0}\right) \cap\left(g(b)+h(b) \cdot \mathbb{Z}_{\geqslant 0}\right)=g(c)+h(c) \cdot \mathbb{Z}_{\geqslant 0} . $$  Then we have $h(c)=\operatorname{lcm}(h(a), h(b))$ since the left hand side must be of the form $m+$ $\operatorname{lcm}(h(a), h(b)) \cdot \mathbb{Z}_{\geqslant 0}$ for some $m$.  If $b$ is a multiple of $a$, then $\operatorname{lcm}(a, b)=b$, so $h(b)=\operatorname{lcm}(h(a), h(b))$, and hence $h(b)$ is a multiple of $h(a)$. Conversely, if $h(b)$ is a multiple of $h(a)$, then $h(b)=\operatorname{lcm}(h(a), h(b))$. On the other hand, we have $h(c)=\operatorname{lcm}(h(a), h(b))$. Since $h$ is injective, we have $c=b$, so $b$ is a multiple of $a$.  We apply the following claim for $H=h$. Claim. Suppose $H: \mathbb{Z}_{\geqslant 2} \rightarrow \mathbb{Z}_{\geqslant 2}$ is a bijection such that $a$ divides $b$ if and only if $H(a)$ divides $H(b)$. Then:  1. $H(p)$ is prime if and only if $p$ is prime; 2. $H\left(\prod_{i=1}^{m} p_{i}^{e_{i}}\right)=\prod_{i=1}^{m} H\left(p_{i}\right)^{e_{i}}$ i.e. $H$ is completely multiplicative; 3. $H$ preserves gcd and lcm.  Proof. We define $H(1)=1$, and consider the bijection $H: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$. By the conditions on $H$, for any $n \in \mathbb{Z}_{\geqslant 2}, n$ and $H(n)$ have the same number of divisors. Hence $H(p)$ is prime if and only if $p$ is prime.  Since the only prime dividing $H\left(p^{r}\right)$ is $H(p)$, we have $H\left(p^{r}\right)=H(p)^{s}$ for some $s \geqslant 1$. Counting the number of divisors, we have $s=r$, so $H\left(p^{r}\right)=H(p)^{r}$ for any prime $p$ and $r \geqslant 1$.  For $a, b \in \mathbb{Z}_{>0}$, recall that $\operatorname{gcd}(a, b)$ is a unique positive integer satisfying the following condition: for any $c \in \mathbb{Z}_{>0}, c$ divides $\operatorname{gcd}(a, b)$ if and only if $c$ divides both $a$ and $b$. By the condition on $H$, for any $c \in \mathbb{Z}_{>0}, H(c)$ divides $H(\operatorname{gcd}(a, b))$ if and only if $H(c)$ divides both $H(a)$ and $H(b)$. Hence we have $H(\operatorname{gcd}(a, b))=\operatorname{gcd}(H(a), H(b))$.  Similarly, we have $H(\operatorname{lcm}(a, b))=\operatorname{lcm}(H(a), H(b))$. Hence we have  $$ \begin{aligned} H\left(\prod_{i=1}^{m} p_{i}^{e_{i}}\right) & =H\left(\operatorname{lcm}\left(p_{1}^{e_{1}}, \ldots, p_{r}^{e_{r}}\right)\right)=\operatorname{lcm}\left(H\left(p_{1}^{e_{1}}\right), \ldots, H\left(p_{r}^{e_{r}}\right)\right)=\operatorname{lcm}\left(H\left(p_{1}\right)^{e_{1}}, \ldots, H\left(p_{r}\right)^{e_{r}}\right) \\ & =\prod_{i=1}^{m} H\left(p_{i}\right)^{e_{i}} \end{aligned} $$  since $H\left(p_{i}\right)$ and $H\left(p_{j}\right)$ are different primes for $i \neq j$. Take two primes $p \neq q$, and let $x, y$ be positive integers such that  $$ g(p)+(x-1) h(p)=g(q)+(y-1) h(q) . $$  This is possible as $h(p)$ and $h(q)$ are two distinct primes. For every $k \geqslant 0$, by $P(p, x+k h(q))$ and $P(q, y+k h(p))$, we have  $$ \left\{\begin{array}{l} g(p \cdot h(x+k h(q)))=g(p)+(x+k h(q)-1) h(p), \\ g(q \cdot h(y+k h(p)))=g(q)+(y+k h(p)-1) h(q), \end{array}\right. $$  where the right hand sides are equal. By the injectivity of $g$, we have  $$ p \cdot h(x+k h(q))=q \cdot h(y+k h(p)) . $$  So, $h(y+k h(p))$ is divisible by $p$ for all $k \geqslant 0$. By the above Claim, $h$ preserves gcd, so  $$ h(\operatorname{gcd}(y, h(p)))=\operatorname{gcd}(h(y), h(y+h(p))) $$  is divisible by $p$. Since $h(p)$ is a prime, $y$ must be divisible by $h(p)$. Moreover, $h(h(p))$ is also a prime, so we have $h(h(p))=p$. The function $h \circ h$ is completely multiplicative, so we have $h(h(n))=n$ for every $n \geqslant 2$.  By $P(a, h(b))$ and $P(b, h(a))$, we have  $$ \left\{\begin{array}{l} g(a b)=g(a \cdot h(h(b))) \stackrel{P(a, h(b))}{=} g(a)+(h(b)-1) h(a), \\ g(b a)=g(b \cdot h(h(a))) \stackrel{P(b, h(a))}{=} g(b)+(h(a)-1) h(b), \end{array}\right. $$  so  $$ g(a)+h(a)(h(b)-1)=g(a b)=g(b)+h(b)(h(a)-1) . $$  Hence $g(a)-h(a)=g(b)-h(b)$ for any $a, b \geqslant 2$, so $g-h$ is a constant function. By comparing the images of $g$ and $h$, the difference is -1 , i.e. $g(a)-h(a)=1$ for any $a \geqslant 2$.  So, we have $g(h(a))=h(h(a))-1=a-1$. By definition,  $$ f(a-1)=h(a)=f^{g(h(a))}(1)=f^{a-1}(1) . $$  By the injectivity of $f$, we have $f^{a-2}(1)=a-1$ for every $a \geqslant 2$. From this, we can deduce inductively that $f(a)=a+1$ for every $a \geqslant 1$. ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-99.jpg?height=293&width=529&top_left_y=2255&top_left_x=769)