Problem 1557

1. Initial Setup and Assumptions:
Given $a, b, c \in \mathbb{Z}^+$ such that $(a^2 - 1, b^2 - 1, c^2 - 1) = 1$, we need to prove that $(ab + c, bc + a, ca + b) = (a, b, c)$.

2. Divisibility in One Direction:
Clearly, $(a, b, c) \mid (ab + c, bc + a, ca + b)$. This is because any common divisor of $a, b, c$ will also divide any linear combination of these terms.

3. **Defining $d$:**
Let $d = (ab + c, bc + a, ca + b)$. We need to show that $d \mid (a, b, c)$.

4. Divisibility by Linear Combinations:
We have:
$$d \mid (a+1)(b+c), \quad d \mid (b+1)(c+a), \quad d \mid (c+1)(a+b)$$
We aim to show that $d \mid (a+b), (b+c), (c+a)$.

5. Prime Divisor Analysis:
Assume $d$ has a prime divisor $p$. Consider $p \mid a+1$. If $p \mid b+1$ and $p \mid c+1$, then $(a+1, b+1, c+1) \geq p > 1$, which contradicts $(a^2 - 1, b^2 - 1, c^2 - 1) = 1$.

6. Case Analysis:
- If $p \mid a+1$ but $p \nmid b+1$ and $p \nmid c+1$, then $p \mid a+b$ and $p \mid a+c$.
- Using $a \equiv b \equiv -1 \pmod{p}$, we get $p = 2$. Thus, $a$ and $b$ are odd, making $c$ even, which contradicts $p \nmid ab + c$.

7. Contradiction and Conclusion:
- If $p \mid a+1$, then $p \nmid b+1$ and $p \nmid c+1$, leading to $p \mid a+b$ and $p \mid a+c$.
- This implies $p \mid (a+1, b-1, c-1)$, another contradiction.

8. No Prime Divisor:
Therefore, no prime $p$ divides $a+1, b+1, c+1$, and thus $d \mid a+b, b+c, c+a$.

9. Further Divisibility:
- $d \mid ab + c \implies d \mid ab - b = a(b-1)$
- Similarly, $d \mid b(c-1)$ and $d \mid c(a-1)$

10. Prime Divisor Analysis Again:
- Assume $p \mid a-1$. Since $(a, a-1) = 1$, $p \mid d \mid a(b-1)$ implies $d \mid b-1$.
- Similarly, $d \mid c-1$, contradicting $(a^2 - 1, b^2 - 1, c^2 - 1) = 1$.

11. Final Conclusion:
No such prime $p$ exists, so $d \mid a, b, c$. Hence, $d \mid (a, b, c)$, and therefore $(ab + c, bc + a, ca + b) \mid (a, b, c)$.

The final answer is $\boxed{ (a, b, c) }$.