Problem 1232

1. Initial Setup and Definitions:
We are given the functional equation:
$$f^{f(a)}(b) f^{f(b)}(a) = (f(a+b))^2$$
for all $a, b \in \mathbb{N}$, where $f^k(n)$ denotes the $k$-th iterate of $f$ applied to $n$.

2. Injectivity and Initial Claims:
- Claim: $f(x) \neq 1$ for all $x$.
- Proof: Suppose $f(k) = 1$ for some $k$. Then, using the functional equation with $a = b = k$:
$$f^{f(k)}(k) f^{f(k)}(k) = (f(2k))^2 \implies f(k) f(k) = (f(2k))^2 \implies 1 = (f(2k))^2$$
This implies $f(2k) = 1$, which contradicts the injectivity of $f$ since $f(k) = f(2k) = 1$. Therefore, $f(x) \neq 1$ for all $x$. $\blacksquare$

- Claim: $f(1) = 2$.
- Proof: Suppose $f(1) \neq 2$. Consider the functional equation with $a = b = 1$:
$$f^{f(1)}(1) f^{f(1)}(1) = (f(2))^2 \implies (f^{f(1)}(1))^2 = (f(2))^2$$
Since $f$ is injective, $f^{f(1)}(1) = f(2)$. If $f(1) = k$, then $f^k(1) = f(2)$. If $k > 2$, this would imply a contradiction as $f$ would not be injective. Therefore, $f(1) = 2$. $\blacksquare$

3. Functional Equation Analysis:
- Using $P(a, 1)$:
$$f^{f(a)}(1) f^2(a) = (f(a+1))^2$$
- Using $P(a, a)$:
$$f^{f(a)}(a) f^{f(a)}(a) = (f(2a))^2 \implies (f^{f(a)}(a))^2 = (f(2a))^2 \implies f^{f(a)}(a) = f(2a)$$

4. Inductive Step:
- Base Case: We have $f(1) = 2$ and $f(2) = 3$.
- Inductive Hypothesis: Assume $f(n) = n+1$ for all $n \leq t$.
- Inductive Step: We need to show $f(t+1) = t+2$.
- Using $P(t, 1)$:
$$f^{f(t)}(1) f^2(t) = (f(t+1))^2$$
Since $f(t) = t+1$, we have:
$$f^{t+1}(1) f^2(t) = (f(t+1))^2$$
Given $f^2(t) = f(f(t)) = f(t+1)$, we get:
$$f^{t+1}(1) f(t+1) = (f(t+1))^2$$
Since $f^{t+1}(1) = t+2$, we have:
$$(t+2) f(t+1) = (f(t+1))^2 \implies t+2 = f(t+1)$$
Thus, $f(t+1) = t+2$.

5. Conclusion:
By induction, we have shown that $f(n) = n+1$ for all $n \in \mathbb{N}$.

The final answer is $\boxed{f(x) = x+1}$.