<s>[INST] 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.
 [/INST] 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
$$x^5-x+2759640=(x^2+12x+377)(x^3-12x^2-233x+7320)$$
</s>