Problem 1629

1. Initial Conditions and Definitions:
- Given $f: \mathbb{N} \to \mathbb{N}$.
- If $a$ and $b$ are coprime, then $f(ab) = f(a)f(b)$.
- For primes $m$ and $k$, $f(m+k-3) = f(m) + f(k) - f(3)$.

2. Base Case:
- Assume $f(1) = 1$.

3. **Define $f(2)$ and $f(4)$:**
- Let $f(2) = a$.
- Let $f(4) = b$.

4. **Using the given functional equation with $m = k = 2$:**
$$f(3) = f(2+2-3) = f(2) + f(2) - f(3) \implies f(3) = 2a - 1$$

5. **Using the given functional equation with $m = 5$ and $k = 2$:**
$$f(5) = f(5+2-3) = f(5) + f(2) - f(3) \implies f(5) = f(4) + f(3) - f(2) = b + (2a - 1) - a = a + b - 1$$

6. **Using the given functional equation with $m = k = 5$:**
$$f(7) = f(5+5-3) = 2f(5) - f(3) \implies f(7) = 2(a + b - 1) - (2a - 1) = 2a + 2b - 2 - 2a + 1 = 2b - 1$$

7. **Using the given functional equation with $m = k = 7$:**
$$f(11) = f(7+7-3) = 2f(7) - f(3) \implies f(11) = 2(2b - 1) - (2a - 1) = 4b - 2 - 2a + 1 = 4b - 2a - 1$$

8. **Using the given functional equation with $m = 11$ and $k = 2$:**
$$f(10) = f(11+2-3) = f(11) + f(2) - f(3) \implies f(10) = 4b - 2a - 1 + a - (2a - 1) = 4b - 3a$$
- Also, since $10 = 2 \times 5$, we have:
$$f(10) = f(2)f(5) = a(a + b - 1)$$
- Equating the two expressions for $f(10)$:
$$4b - 3a = a(a + b - 1)$$

9. **Using the given functional equation with $m = 13$ and $k = 2$:**
$$f(12) = f(13+2-3) = f(13) + f(2) - f(3) \implies f(12) = 5b - 3a - 1 + a - (2a - 1) = 5b - 4a$$
- Also, since $12 = 3 \times 4$, we have:
$$f(12) = f(3)f(4) = (2a - 1)b$$
- Equating the two expressions for $f(12)$:
$$5b - 4a = b(2a - 1)$$

10. Solving the system of equations:
- From $4b - 3a = a(a + b - 1)$:
$$4b - 3a = a^2 + ab - a$$
- From $5b - 4a = b(2a - 1)$:
$$5b - 4a = 2ab - b$$
- Simplifying the second equation:
$$5b - 4a = 2ab - b \implies 5b - b = 2ab - 4a \implies 4b = 2ab - 4a \implies 4b = 2a(b - 2) \implies b = \frac{2a}{3 - a}$$
- Substituting $b = \frac{2a}{3 - a}$ into the first equation:
$$4\left(\frac{2a}{3 - a}\right) - 3a = a^2 + a\left(\frac{2a}{3 - a}\right) - a$$
$$\frac{8a}{3 - a} - 3a = a^2 + \frac{2a^2}{3 - a} - a$$
$$\frac{8a - 3a(3 - a)}{3 - a} = a^2 + \frac{2a^2 - a(3 - a)}{3 - a}$$
$$\frac{8a - 9a + 3a^2}{3 - a} = a^2 + \frac{2a^2 - 3a + a^2}{3 - a}$$
$$\frac{-a + 3a^2}{3 - a} = a^2 + \frac{3a^2 - 3a}{3 - a}$$
$$-a + 3a^2 = 3a^2 - 3a$$
$$-a = -3a$$
$$2a = 0 \implies a = 1$$
- Substituting $a = 1$ into $b = \frac{2a}{3 - a}$:
$$b = \frac{2 \cdot 1}{3 - 1} = \frac{2}{2} = 1$$
- Therefore, $(a, b) = (1, 1)$.

11. Checking the other solution:
- Solving $a^2 - 3a + 2 = 0$:
$$(a - 1)(a - 2) = 0 \implies a = 1 \text{ or } a = 2$$
- If $a = 2$:
$$b = \frac{2 \cdot 2}{3 - 2} = \frac{4}{1} = 4$$
- Therefore, $(a, b) = (2, 4)$.

12. **Finding $f(11)$:**
- For $(a, b) = (1, 1)$:
$$f(11) = 4b - 2a - 1 = 4 \cdot 1 - 2 \cdot 1 - 1 = 4 - 2 - 1 = 1$$
- For $(a, b) = (2, 4)$:
$$f(11) = 4b - 2a - 1 = 4 \cdot 4 - 2 \cdot 2 - 1 = 16 - 4 - 1 = 11$$

Conclusion:
- Both values are possible for functions $f(n) = 1$ or $f(n) = n$.

The final answer is $\boxed{1 \text{ or } 11}$