<s>[INST] Factor $x^5-x+15$ It's possible to factor $x^5-x+15$. WolframAlpha gives the answer of: $$(x^2+x+3)(x^3-x^2-2x+5)$$
According to the wikipedia article on quintic functions, the general form $x^5-x+a$ is factorable only when $a=±15$, $±22440$, or $±2759640$.
Question: How would one factor such an expression? For me, it seems close to impossible by hand.
Bonus question: Why do those specific values of $a$ make the expression factorable? If there's no simple answer, is there a paper/further reading about it?
 [/INST] One can factor $$f(x) := x^5 - x \pm 15$$ manually (over $\Bbb Z$) without too much fuss.
First, if $f$ has a linear factor, it has a rational root and (because $f$ is monic) any rational root must be an integer. But $f(0) \equiv f(1) \equiv 1 \pmod 2$, so $f$ has no root modulo $2$ and hence no integer root and hence no linear factor. (Alternatively, we can show this with the Rational Root Theorem; see the bottom of this answer.)
Thus, if $f$ factors, it must factor as a product of a cubic and a quadratic, that is, (where we denote $\Lambda = \pm 15$)
$$x^5 - x + \Lambda = (x^3 + A x^2 + B x + C) (x^2 + D x + E)$$
for some integers $A, B, C, D, E$. Distributing the right-hand side and comparing coefficients gives various (at most) quadratic conditions on those integers:
\begin{align}
A + D &= 0 \\
A D + B + E &= 0 \\
A E + B D + C &= 0 \\
B E + C D &= - 1 \\
C E &= \Lambda \\
\end{align}
We can quickly reduce this system: The first equation gives $D = -A$, so the second equation gives $E = A^2 - B$, and then the third equation becomes $C = -A^3 + 2 A B$. Substituting leaves the system
\begin{align*}
A^4 - A^2 B - B^2 &= -1 \\
A (A^2 - 2 B) (A^2 - B) &= \Lambda .
\end{align*}
Since $A, A^2 - 2 B, A^2 - B$ are all integers and the prime factorization of $|\Lambda| = 15$ is $3 \cdot 5$, there are only a small number of combinations to check (in fact, since $15$ is a product of two primes, one of the three factors must be $\pm 1$), and we can quickly recover the factorization mentioned in the question.
Remark For readers unfamiliar with reducing polynomial equations modulo primes, we can conclude that $f$ has no linear factor with just a little more work, using the Rational Root Theorem, which in this case implies that the only possible rational roots are $\pm 1, \pm 3, \pm 5, \pm 15$; substituting shows that none of these are roots. One can even avoid most of this work: The derivative of the polynomial is $f'(x) = 5 x^4 - 1$, so $f$ is increasing where $|x| \geq 1$, but $f(-3) = -69, f(-1) = 13, f(1) = 15$, from which we can conclude none of the eight candidates are roots.
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