Find all $a, b, c, d$ Find all $a, b, c, d$ satisfying
$$\frac{x^4+ax^3+bx^2-8x+4}{(x^2+cx+d)^2} = 1$$
My answers are: 
$$\begin{cases}
d=2\\
d=-2
\end{cases}
\quad \begin{cases}
a=4\\
a=-4
\end{cases} \quad
\begin{cases}
c=2\\
c=-2
\end{cases} \quad
\begin{cases}
b=8\\
b=0
\end{cases}$$ 
Is it right?

For another perspective:
Let $r_1, r_2, r_3, r_4$ be the zeroes of
$$f(x) = x^4 + ax^3 + bx^2 - 8x + 4.$$
Since 
$$x^4 + ax^3 + bx^2 - 8x + 4 = (x^2 + cx + d)^2,$$ 
$r_1, r_2, r_3, r_4$ are also the zeroes of
$$g(x) = (x^2 + cx + d)^2.$$
Using the quadratic formula, we get (without loss of generality) that
$$r_1 = r_2 = \dfrac{-c - \sqrt{c^2 - 4d}}{2}$$
and
$$r_3 = r_4 = \dfrac{-c + \sqrt{c^2 - 4d}}{2}.$$
Now, we have the equations
$$4 = {r_1}\cdot{r_2}\cdot{r_3}\cdot{r_4}$$
$$-(-8) = {r_1}{r_2}{r_3} + {r_1}{r_2}{r_4} + {r_1}{r_3}{r_4} + {r_2}{r_3}{r_4}$$
$$b = {r_1}{r_2} + {r_1}{r_3} + {r_1}{r_4} + {r_2}{r_3} + {r_2}{r_4} + {r_3}{r_4}$$
$$-a = r_1 + r_2 + r_3 + r_4,$$
which can then be solved for $a, b, c, d$.
In particular, from
$$4 = {r_1}\cdot{r_2}\cdot{r_3}\cdot{r_4}$$
we get
$$4 = \left(\dfrac{c^2 - (c^2 - 4d)}{4}\right)^2 = \left(\dfrac{4d}{4}\right)^2 \iff 4 = d^2 \iff d = \pm 2.$$
I leave the other computations as an exercise to the interested reader.  =)
