4. In the Cartesian coordinate system $x O y$, the curve $y=x^{3}-a x$ has two parallel tangent lines. If the slopes of these two tangent lines are both 1, and the distance between the two tangent lines is 8, then the value of the real number $a$ is $\qquad$.
Solving
$y^{\prime}=3 x^{2}-a=1 \Rightarrow 3 x^{2}=a+1 \Rightarrow x_{1,2}= \pm \sqrt{\frac{a+1}{3}}$, then the equation of the tangent line passing through $\left(x_{1}, y_{1}\right)$ is
$$
y-\left(x_{1}^{3}-a x_{1}\right)=x-x_{1} \Rightarrow y=x+x_{1}^{3}-(a+1) x_{1} \Rightarrow y=x-\frac{2(a+1)}{3} x_{1} \text {, }
$$

Similarly, the equation of the other tangent line is $y=x-\frac{2(a+1)}{3} x_{2}$. According to the problem,
$$
\frac{2(a+1)}{3}\left(x_{1}-x_{2}\right)=\frac{4(a+1)}{3} \sqrt{\frac{a+1}{3}}=8 \sqrt{2} \Rightarrow \frac{a+1}{3} \sqrt{\frac{a+1}{3}}=2 \sqrt{2} \Rightarrow \frac{a+1}{3}=2 \Rightarrow a=5 .
$$