Problem 1671

### Part (a)
1. Given $c = 0$, we have $a_n = n^2$ and $a_{n+1} = (n+1)^2$.
2. We need to find $d_n = \gcd(a_n, a_{n+1}) = \gcd(n^2, (n+1)^2)$.
3. Using the property of gcd, $\gcd(a, b) = \gcd(a, b - a)$, we get:
$$\gcd(n^2, (n+1)^2) = \gcd(n^2, (n+1)^2 - n^2) = \gcd(n^2, 2n + 1)$$
4. Since $n^2$ and $2n + 1$ are coprime (i.e., $\gcd(n, n+1) = 1$), it follows that:
$$\gcd(n^2, 2n + 1) = 1$$

Conclusion:
$$d_n = 1 \quad \forall n \geq 1$$

### Part (b)
1. Given $c = 1$, we have $a_n = n^2 + 1$ and $a_{n+1} = (n+1)^2 + 1 = n^2 + 2n + 2$.
2. We need to find $d_n = \gcd(a_n, a_{n+1}) = \gcd(n^2 + 1, n^2 + 2n + 2)$.
3. Using the Euclidean Algorithm:
$$\gcd(n^2 + 1, n^2 + 2n + 2) = \gcd(n^2 + 1, (n^2 + 2n + 2) - (n^2 + 1)) = \gcd(n^2 + 1, 2n + 1)$$
4. Consider $n$ even, $n = 2k$:
$$\gcd(4k^2 + 1, 4k + 1)$$
Since $4k^2 + 1 \equiv 1 \pmod{4k + 1}$, we have:
$$\gcd(4k^2 + 1, 4k + 1) = \gcd(1, 4k + 1) = 1$$
5. Consider $n$ odd, $n = 2k - 1$:
$$\gcd(4k^2 - 4k + 2, 4k^2 + 1)$$
Since $4k^2 + 1 \equiv 1 \pmod{4k - 1}$, we have:
$$\gcd(4k^2 + 1, 4k - 1) = \gcd(1, 4k - 1) = 1$$
6. Therefore, $d_n \in \{1, 5\}$.

Conclusion:
$$d_n \in \{1, 5\} \quad \forall n \geq 1$$

### Part (c)
1. Given $a_n = n^2 + c$ and $a_{n+1} = (n+1)^2 + c = n^2 + 2n + 1 + c$.
2. We need to find $d_n = \gcd(a_n, a_{n+1}) = \gcd(n^2 + c, n^2 + 2n + 1 + c)$.
3. Using the Euclidean Algorithm:
$$\gcd(n^2 + c, n^2 + 2n + 1 + c) = \gcd(n^2 + c, (n^2 + 2n + 1 + c) - (n^2 + c)) = \gcd(n^2 + c, 2n + 1)$$
4. Consider $n$ even, $n = 2k$:
$$\gcd(4k^2 + c, 4k + 1)$$
Since $4k^2 + c \equiv c \pmod{4k + 1}$, we have:
$$\gcd(4k^2 + c, 4k + 1) = \gcd(c, 4k + 1)$$
5. Consider $n$ odd, $n = 2k - 1$:
$$\gcd(4k^2 - 4k + 1 + c, 4k^2 + c)$$
Since $4k^2 + c \equiv c \pmod{4k - 1}$, we have:
$$\gcd(4k^2 + c, 4k - 1) = \gcd(c, 4k - 1)$$
6. Therefore, $d_n \leq 4c + 1$.

Conclusion:
$$d_n \leq 4c + 1 \quad \forall n \geq 1$$

The final answer is $\boxed{ d_n = 1 }$ for part (a), $d_n \in \{1, 5\}$ for part (b), and $d_n \leq 4c + 1$ for part (c).