What is the product of real numbers $a$ which make $x^2+ax+1$ a negative integer for only one real number $x$?

$ 
\textbf{(A)}\ -1
\qquad\textbf{(B)}\ -2
\qquad\textbf{(C)}\ -4
\qquad\textbf{(D)}\ -6
\qquad\textbf{(E)}\ -8
$
1. We start with the quadratic equation \( f(x) = x^2 + ax + 1 \). We need to find the values of \( a \) such that \( f(x) \) is a negative integer for only one real number \( x \).

2. For a quadratic function \( f(x) = x^2 + ax + 1 \) to be a negative integer at only one point, it must have a minimum value that is a negative integer. The vertex of the parabola \( f(x) \) is given by \( x = -\frac{a}{2} \).

3. The value of the function at the vertex is:
   \[
   f\left(-\frac{a}{2}\right) = \left(-\frac{a}{2}\right)^2 + a\left(-\frac{a}{2}\right) + 1 = \frac{a^2}{4} - \frac{a^2}{2} + 1 = 1 - \frac{a^2}{4}
   \]

4. We need this minimum value to be a negative integer. Let’s denote this negative integer by \( -k \), where \( k \) is a positive integer. Thus, we have:
   \[
   1 - \frac{a^2}{4} = -k
   \]

5. Solving for \( a^2 \), we get:
   \[
   1 + k = \frac{a^2}{4}
   \]
   \[
   a^2 = 4(1 + k)
   \]

6. Since \( a \) must be real, \( 1 + k \) must be positive. For the quadratic to have a minimum value that is a negative integer for only one real number \( x \), \( k \) must be 1 (the smallest positive integer). Thus:
   \[
   a^2 = 4(1 + 1) = 4 \cdot 2 = 8
   \]
   \[
   a = \pm \sqrt{8} = \pm 2\sqrt{2}
   \]

7. The product of the possible values of \( a \) is:
   \[
   (2\sqrt{2})(-2\sqrt{2}) = -8
   \]

Conclusion:
\[
\boxed{-8}
\]