Find all functions $f:\mathbb{N} \rightarrow \mathbb{N}$, such that $f(a)f(a+b)-ab$ is a perfect square for all $a, b \in \mathbb{N}$. 
To solve the problem, we need to find all functions \( f: \mathbb{N} \rightarrow \mathbb{N} \) such that \( f(a)f(a+b) - ab \) is a perfect square for all \( a, b \in \mathbb{N} \).

1. **Initial Assumption and Lemma:**
   - Assume \( p \) is an odd prime such that \( p \mid f(a) \). Then \( f(a)f(a+b) - ab \) must be a perfect square.
   - If \( p \nmid a \), then \( -ab \) generates the complete residue system \( \{0, 1, \ldots, p-1\} \) as we iterate over \( b \).
   - The set of quadratic residues modulo \( p \) is \( \{0^2, 1^2, \ldots, \left(\frac{p-1}{2}\right)^2\} \), which contains exactly \( \frac{p-1}{2} \) non-quadratic residues.
   - Since \( p \geq 3 \), there exists \( c \in \mathbb{Z}/p\mathbb{Z} \) which is a non-quadratic residue. Taking \( b = -c \cdot a^{-1} \) in \( \mathbb{Z}/p\mathbb{Z} \) generates a contradiction.

   **Lemma:** If \( p \mid f(a) \) where \( p \) is an odd prime, then \( p \mid a \).

2. **Odd and Even Analysis:**
   - Prove that if \( x \) is odd, then \( f(x) \) is also odd.
   - By contradiction, let \( t \) be odd and \( 2 \mid f(t) \).
   - If \( 4 \mid f(t) \), then \( f(t)f(t+1) - t \) and \( f(t)f(t+3) - 3t \) are congruent to \( -t \) and \( -3t \) modulo 4, respectively. One of these residues is congruent to 3 modulo 4, which is a non-quadratic residue in \( \mathbb{Z}/4\mathbb{Z} \).

3. **Case Analysis for \( f(p) \):**
   - **Case 1:** \( f(p) = 2 \cdot p^m \) where \( m \) is non-negative.
     - If \( m \geq 2 \), then \( f(p)f(p+1) - p \) is a square divisible by \( p \). Therefore, \( p^2 \mid f(p)f(p+1) - p \), but since \( p^2 \mid f(p) \), we end up with \( p \mid 1 \), a contradiction.
     - If \( m = 1 \), then \( f(p) = 2p \).
     - Further analysis shows contradictions when considering \( f(2p) \) and \( f(3p) \).

   - **Case 2:** \( f(p) = p^m \) where \( m \) is non-negative.
     - If \( m \geq 2 \), then \( f(p)f(p+1) - p \) is a square divisible by \( p \). Therefore, \( p^2 \mid f(p)f(p+1) - p \), but since \( p^2 \mid f(p) \), we end up with \( p \mid 1 \), a contradiction.
     - Therefore, \( f(p) \) is either 1 or \( p \) for all primes \( p > 2 \).

4. **Final Conclusion:**
   - Suppose there exists a prime \( r > 2 \) with \( f(r) = 1 \). Choose any prime \( s > r^2 > 2 \). Then \( f(r)f(s) - r(s-r) \) is a square, so \( r(s-r) \leq f(r)f(s) = f(s) \leq s \). Thus, \( r^2 \geq s \cdot (r-1) \geq s > r^2 \), a contradiction.
   - Therefore, \( f(p) = p \) for all primes \( p > 2 \).

5. **Infinite Sequence of Fixed Points:**
   - If we have an infinite sequence \( y_n \) such that \( f(y_n) = y_n \), then for any natural \( x \), there exists \( z \geq 1 \) such that \( y_z > x^2 > x \).
   - By contradiction, if \( f(x) < x \), then \( f(x) \leq x-1 \). Since \( f(x)f(y_z) - x(y_z - x) \) is a square, \( x \cdot (y_z - x) \leq f(x)f(y_z) \leq (x-1) \cdot y_z \), implying \( y_z \leq x^2 \), a contradiction.
   - Therefore, all odd primes are fixed points, and we conclude that \( f(x) = x \) for all \( x \).

\(\blacksquare\)

The final answer is \( \boxed{ f(x) = x } \) for all \( x \in \mathbb{N} \).