Values of $a$ such that $x^5-x-a$ has quadratic factor I would like to find all integers $a$ such that $x^5-x-a$ has a quadratic factor in $\mathbb{Z}[x]$.
My Attempt
Let $x^5-x-a=(x^2+bx+c)(x^3+dx^2+ex+f)$, so that we have the following:
$$\begin{array}{rcl}
b+d&=&0\\
e+bd+c&=&0\\
f+be+cd&=&0\\
bf+ce&=&-1\\
cf&=&-a
\end{array}$$
Hence:
$$\begin{array}{rcccl}
d&=&-b\\
e&=&-bd-c&=&b^2-c\\
f&=&-be-cd&=&-b^3+2bc
\end{array}$$
and we have:
$$1=-bf-ce=b^4-3b^2c+c^2,$$
so that:
$$(2c-3b^2)^2=5b^4+4.$$
Question
How can I find all values of $n$ such that $5n^4+4$ is a perfect square?
My Attempt
If $m^2=5n^4+4$, then $m^2-5n^4=4$.
If $m=2m_*$, then $n$ is even, so that $n=2n_*$, and we have the equation $m_*^2-20n_*^4=1$. By Pell equation, since $(a,b)=(9,2)$ is the least non-trivial solution of $a^2-20b^2=1$, then the general solution has the form $(a_n,b_n)$ where $a_n+b_n\sqrt{20}=(9+2\sqrt{20})^n$, but I do not know how to find out what values of $n$ make $b_n$ a square.
 A: This is essentially an elliptic curve.
There might be elementary methods, but there are also computer algebra systems that can (in many cases) solve this kind of diophantine equations.
We may rewrite the equation as: $m^2n^2 = 5n^6 + 4n^2$.
If we write $y = 5mn$ and $x = 5n^2$, then it becomes $y^2 = x^3 + 20x$.
Now we use Sage to find all integer points on this curve. Paste the following codes into this site and press "Evaluate".
EllipticCurve([20, 0]).integral_points()

The output:
[(0 : 0 : 1), (4 : 12 : 1), (5 : 15 : 1), (720 : 19320 : 1)]

We see that the corresponding values of $(m, n)$ are $(2,0), (3,1), (322,12)$, respectively (negative values are not listed).
A: Another way is to do a long division of $x ^ 5-x-a$ by the arbitrary trinomial $x ^ 2 + bx + c$ and set the remainder to zero. This gives the remainder
$$(c ^ 2-3b ^ 2c + b ^ 4-1) x + (cb ^ 3-2bc ^ 2-a) = 0$$ from where we have
$a = cb (b ^ 2-2c)$ and $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$.
This means that for every solution of $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$ we have a corresponding value $a = cb (b ^ 2-2c)$.
Some solutions of $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$ are $(b,c)=(1,3),(0,1),(12,55),(12,377)$.
EXAMPLES.-$(b,c)=(1,3)$ gives $a=-15$ and we have
$$x^5-x+15=(x^2+x+3)(x^3-x^2-2x+5)$$
$(b,c)=(12,377)$ gives $a=-2759640$ and we have
$$x^5-x+2759640=(x^2+12x+377)(x^3-12x^2-233x+7320)$$
