A $\textit{cubic sequence}$ is a sequence of integers given by $a_n =n^3 + bn^2 + cn + d$, where $b, c$ and $d$ are integer constants and $n$ ranges over all integers, including negative integers.
$\textbf{(a)}$ Show that there exists a cubic sequence such that the only terms
of the sequence which are squares of integers are $a_{2015}$ and $a_{2016}$.
$\textbf{(b)}$ Determine the possible values of $a_{2015} \cdot a_{2016}$ for a cubic sequence
satisfying the condition in part $\textbf{(a)}$.

1. **Shifting the Sequence:**
   We start by shifting the sequence such that \( a_{2015} \mapsto a_0 \) and \( a_{2016} \mapsto a_1 \). This simplifies our problem to finding a cubic polynomial \( a_n = n^3 + bn^2 + cn + d \) such that \( a_0 \) and \( a_1 \) are perfect squares, and no other terms in the sequence are perfect squares.

2. **Formulating the Polynomial:**
   Consider the polynomial \( y = (rx + s)^2 \). We set \( a_0 = s^2 \) and \( a_1 = (r + s)^2 \). This gives us the polynomial:
   \[
   y = x^3 + bx^2 + cx + s^2 - (rx + s)^2 = x^3 + (b - r^2)x^2 + (c - 2rs)x
   \]
   We need this polynomial to have a double root, which implies that the discriminant of the quadratic part must be zero.

3. **Finding the Roots:**
   Transforming our polynomial, we get:
   \[
   y = x^3 + (b - r^2)x^2 + (r^2 - b - 1)x
   \]
   The roots of this polynomial are \( 0, 1, r^2 - b - 1 \). For \( r^2 - b - 1 \) to be a root, we need \( r^2 = b + 1 \) or \( r^2 = b + 2 \).

4. **Constructing the Polynomial:**
   This means our difference polynomials are:
   \[
   y = x^3 - x^2 \quad \text{or} \quad y = x^3 - 2x^2 + x
   \]
   Therefore, our overall polynomials are:
   \[
   y = (rx + s)^2 + x^3 - x^2 \quad \text{or} \quad y = (rx + s)^2 + x^3 - 2x^2 + x
   \]

5. **Selecting Values:**
   Selecting \( x = 4s(r + s) \) in the first polynomial and \( x = 4s(r + s) + 1 \) in the second polynomial, we get:
   \[
   y = s^2(4(r + s)(r + 2s) - 1)^2 \quad \text{and} \quad y = (4r^2s + 12rs^2 + r + 8s^3 + s)^2
   \]
   This forces one of \( s \) or \( r + s \) to be zero.

6. **Conclusion:**
   Hence, the product is always \( \boxed{0} \). Finally, we claim that \( a_n = n^3 - n^2 + n \) is a valid cubic sequence. Indeed, note that \( a_n \) factors as \( n(n^2 - n + 1) \), and since these two factors are coprime, it follows that both must attain perfect square values. It is easy to see by bounding that \( n^2 - n + 1 \) only does so for \( n = 0, 1 \), whence the conclusion follows.