Determine all the injective functions $f : \mathbb{Z}_+ \to \mathbb{Z}_+$, such that for each pair of integers $(m, n)$ the following conditions hold:

$a)$ $f(mn) = f(m)f(n)$
$b)$ $f(m^2 + n^2) \mid  f(m^2) +  f(n^2).$
1. **Initial Setup and Definitions**:
   - Let \( f : \mathbb{Z}_+ \to \mathbb{Z}_+ \) be an injective function.
   - We are given two conditions:
     \[
     a) \quad f(mn) = f(m)f(n)
     \]
     \[
     b) \quad f(m^2 + n^2) \mid f(m^2) + f(n^2)
     \]
   - Let \( P(m, n) \) denote the statement \( f(m^2 + n^2) \mid f(m^2) + f(n^2) \).

2. **Multiplicativity and Initial Value**:
   - From the multiplicativity condition \( f(mn) = f(m)f(n) \), we can deduce that \( f(1) = 1 \) because:
     \[
     f(1 \cdot 1) = f(1)f(1) \implies f(1) = 1
     \]

3. **Claim**: \( f(p) = p \) for all primes \( p \equiv 1 \pmod{4} \).

4. **Proof of the Claim**:
   - Since \( p \equiv 1 \pmod{4} \), there exist integers \( m, n > 0 \) such that \( p = m^2 + n^2 \).
   - Therefore, \( f(p) \mid f(m^2) + f(n^2) = f(m)^2 + f(n)^2 \).
   - By the multiplicativity of \( f \), we have:
     \[
     f(p) \mid f(m^2) + f(n^2) = f(m)^2 + f(n)^2
     \]
   - Using LTE (Lifting The Exponent lemma), for any \( k \geq 1 \):
     \[
     p^{k+1} \mid m^{2p^k} + n^{2p^k}
     \]
   - Applying \( P(m^{p^k}, n^{p^k}) \) and multiplicativity:
     \[
     f(p)^{k+1} \mid f(m^{2p^k} + n^{2p^k}) \mid f(m^2)^{p^k} + f(n^2)^{p^k}
     \]
   - If some prime \( q \neq p \) divides \( f(p) \), then by LTE:
     \[
     (1+k)v_q(f(p)) \leq v_q(f(m)^2 + f(n)^2)
     \]
     which is impossible for large \( k \). Therefore, \( f(p) \) is a power of \( p \).

5. **Conclusion of the Claim**:
   - Since \( f \) is injective and \( f(p) > 1 \), we have \( p \mid f(p) \).
   - By LTE, \( (1+k)v_p(f(p)) \leq v_p(f(m)^2 + f(n)^2) + k \), which is only possible if \( v_p(f(p)) = 1 \), i.e., \( f(p) = p \).

6. **Extension to All Positive Integers**:
   - If \( m, n \) are coprime positive integers with different parities, the only prime factors of \( m^2 + n^2 \) are \( \equiv 1 \pmod{4} \).
   - By the claim and multiplicativity, \( f(m^2 + n^2) = m^2 + n^2 \).
   - Fix \( n \) and let \( p \) be a sufficiently large prime \( \equiv 1 \pmod{4} \):
     \[
     f(p^2 + 4n^2) = p^2 + 4n^2
     \]
   - Since \( f(p^2 + 4n^2) \mid f(p)^2 + f(2n)^2 = p^2 + f(2n)^2 \):
     \[
     p^2 + 4n^2 \mid f(2n)^2 - 4n^2
     \]
   - Taking \( p \) large enough, we get \( f(2n) = 2n \) for all \( n \geq 1 \).

7. **Final Conclusion**:
   - Taking \( n = 1 \) gives \( f(2) = 2 \).
   - By multiplicativity, \( f(n) = n \) for all \( n \in \mathbb{Z}_+ \).

\(\boxed{f(n) = n \ \ \forall n \in \mathbb{Z}_+}\)