Find all sets of natural numbers $(a, b, c)$ such that $$a+1|b^2+c^2\,\, , b+1|c^2+a^2\,\,, c+1|a^2+b^2.$$
To find all sets of natural numbers \((a, b, c)\) such that 
\[ a+1 \mid b^2 + c^2, \quad b+1 \mid c^2 + a^2, \quad c+1 \mid a^2 + b^2, \]
we will analyze the given conditions step by step.

1. **Analyze the divisibility conditions:**
   - We need \(a+1\) to divide \(b^2 + c^2\).
   - We need \(b+1\) to divide \(c^2 + a^2\).
   - We need \(c+1\) to divide \(a^2 + b^2\).

2. **Consider the example provided:**
   - Let \(a = k\), \(b = k^2 - 1\), and \(c = k^2(k^2 - 1)\) for some natural number \(k\).

3. **Verify the example:**
   - Check if \(a+1 \mid b^2 + c^2\):
     \[
     a = k \implies a+1 = k+1
     \]
     \[
     b = k^2 - 1 \implies b^2 = (k^2 - 1)^2 = k^4 - 2k^2 + 1
     \]
     \[
     c = k^2(k^2 - 1) \implies c^2 = [k^2(k^2 - 1)]^2 = k^4(k^2 - 1)^2 = k^8 - 2k^6 + k^4
     \]
     \[
     b^2 + c^2 = k^4 - 2k^2 + 1 + k^8 - 2k^6 + k^4 = k^8 - 2k^6 + 2k^4 - 2k^2 + 1
     \]
     We need to check if \(k+1\) divides \(k^8 - 2k^6 + 2k^4 - 2k^2 + 1\).

4. **Simplify the expression modulo \(k+1\):**
   - Note that \(k \equiv -1 \pmod{k+1}\):
     \[
     k^2 \equiv 1 \pmod{k+1}
     \]
     \[
     k^4 \equiv 1 \pmod{k+1}
     \]
     \[
     k^6 \equiv 1 \pmod{k+1}
     \]
     \[
     k^8 \equiv 1 \pmod{k+1}
     \]
     Substituting these into the expression:
     \[
     k^8 - 2k^6 + 2k^4 - 2k^2 + 1 \equiv 1 - 2 \cdot 1 + 2 \cdot 1 - 2 \cdot 1 + 1 \equiv 1 - 2 + 2 - 2 + 1 \equiv 0 \pmod{k+1}
     \]
     Thus, \(k+1 \mid b^2 + c^2\).

5. **Check the other conditions similarly:**
   - For \(b+1 \mid c^2 + a^2\):
     \[
     b = k^2 - 1 \implies b+1 = k^2
     \]
     \[
     c = k^2(k^2 - 1) \implies c^2 = k^8 - 2k^6 + k^4
     \]
     \[
     a = k \implies a^2 = k^2
     \]
     \[
     c^2 + a^2 = k^8 - 2k^6 + k^4 + k^2
     \]
     We need to check if \(k^2\) divides \(k^8 - 2k^6 + k^4 + k^2\):
     \[
     k^8 - 2k^6 + k^4 + k^2 = k^2(k^6 - 2k^4 + k^2 + 1)
     \]
     Clearly, \(k^2 \mid k^2(k^6 - 2k^4 + k^2 + 1)\).

   - For \(c+1 \mid a^2 + b^2\):
     \[
     c = k^2(k^2 - 1) \implies c+1 = k^2(k^2 - 1) + 1
     \]
     \[
     a = k \implies a^2 = k^2
     \]
     \[
     b = k^2 - 1 \implies b^2 = k^4 - 2k^2 + 1
     \]
     \[
     a^2 + b^2 = k^2 + k^4 - 2k^2 + 1 = k^4 - k^2 + 1
     \]
     We need to check if \(k^2(k^2 - 1) + 1\) divides \(k^4 - k^2 + 1\).

6. **Conclusion:**
   - The example provided satisfies all the conditions, and thus there are infinitely many solutions of the form \((a, b, c) = (k, k^2 - 1, k^2(k^2 - 1))\).

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