Find all the natural numbers $a,b,c$ such that:

1) $a^2+1$ and $b^2+1$ are primes
2) $(a^2+1)(b^2+1)=(c^2+1)$
To find all natural numbers \(a, b, c\) such that:
1. \(a^2 + 1\) and \(b^2 + 1\) are primes.
2. \((a^2 + 1)(b^2 + 1) = c^2 + 1\),

we start by analyzing the given conditions.

1. **Assume \(a \geq b\)**:
   - This assumption helps us to simplify the problem by considering the larger value first.

2. **Bound for \(c\)**:
   - Since \((a^2 + 1)(b^2 + 1) = c^2 + 1\), we have:
     \[
     c^2 + 1 \leq (a^2 + 1)^2
     \]
   - This implies:
     \[
     c^2 \leq (a^2 + 1)^2 - 1
     \]
   - Therefore:
     \[
     c < a^2 + 1
     \]

3. **Divisibility condition**:
   - We know that \(a^2 + 1\) divides \((c^2 + 1) - (a^2 + 1)\):
     \[
     a^2 + 1 \mid (c^2 + 1) - (a^2 + 1) = c^2 - a^2 = (c - a)(c + a)
     \]

4. **Range for \((c - a)(c + a)\)**:
   - Since \(0 < c - a < c + a < a^2 + a + 1 < 2(a^2 + 1)\), the only possible values for \((c - a)(c + a)\) are \(a^2 + 1\) or \(1\).

5. **Case analysis**:
   - If \(a^2 + 1 = c + a\) and \(c - a = 1\):
     - Solving these equations, we get:
       \[
       c = a + 1
       \]
       \[
       a^2 + 1 = (a + 1) + a = 2a + 1
       \]
     - This simplifies to:
       \[
       a^2 - 2a = 0 \implies a(a - 2) = 0
       \]
     - Since \(a\) is a natural number, \(a = 2\).

6. **Finding \(b\)**:
   - If \(a = 2\), then \(a^2 + 1 = 5\), which is a prime.
   - We need to find \(b\) such that \(b^2 + 1\) is also a prime and \((5)(b^2 + 1) = c^2 + 1\).
   - Trying \(b = 1\):
     \[
     b^2 + 1 = 2 \quad \text{(which is a prime)}
     \]
     \[
     (5)(2) = 10 = c^2 + 1 \implies c^2 = 9 \implies c = 3
     \]

Thus, the only solution is \(a = 2\), \(b = 1\), and \(c = 3\).

The final answer is \( \boxed{ (a, b, c) = (2, 1, 3) } \).