x(x+1)(x+a+1)(x+a) =a²
x(x+a+1) x (x+1)(x+a) = a²
[x² + (a+1)x] x [x² + (a+1)x + a] = a²

Let x² + (a+1)x = z

z(z+a) = a²
This equation has a non-negative discriminant. So this equation has real roots irrespective of the values of a.
z = (-a/2)(-15)

Now,
x² + (a+1)x = (-a/2)(-1+5)
x² + (a+1)x + (a/2)(-1+5) = 0
For this equation to have real roots, its discriminant must be non-zero
(a+1)² - a²(-1+5)² 0
[a - 1/(5 - 2)][a + 1/5] 0
a ( - , -1/5] [1/(5 - 2), ) ----------------------(1)

And,
x² + (a+1)x = (-a/2)(-1-5)
x² + (a+1)x - (a/2)(1+5) = 0
For this equation to have real roots, its discriminant must be non-zero
(a+1)² - a²(-1+5)² 0
[a + 1/(5 + 2)][a - 1/5] 0
a [-1/(5 + 2), 1/5] ----------------------(2)

For four real roots, (1) and (2) must be simultaneously true, which isn't possible. Hence there is no such value of a for which the given equation has 4 real roots.