TITLE: Factor $x^5-x+15$
QUESTION [11 upvotes]: 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?

REPLY [3 votes]: Wednesday morning. I wanted to see what happened with my Fibonacci trick for the largest constant term. I now know that the list of successful constant terms is a minor consequence of the fact that 144 is the largest square Fibonacci number. This was proved by J. H. E. Cohn in 1964.
Square Fibonacci numbers
Summary of proof on square Fibonacci numbers
There is no integer root to $x^5 - x \pm 2759640.$  We arrive at
$$ (x^3 + A x^2 + B x + C)(x^2 + D x + E) = x^5 - x \pm 2759640 $$
The point is not to solve the whole system at once, rather do one coefficient at a time and rewrite the system.  The degree four term must be 0, so $A+D = 0$
$$ (x^3 + A x^2 + B x + C)(x^2 -A x + E) = x^5 - x \pm 2759640 $$
Next the cubed term is zero, so $E-A^2 + B = 0, $ or $E = A^2 - B.$
$$ (x^3 + A x^2 + B x + C)(x^2 -A x + (A^2 - B)) = x^5 - x \pm 2759640 $$
Next $x^2$ has 0, or $A^3 - AB -AB + C = 0,$ or $C = 2AB - A^3,$
$$ (x^3 + A x^2 + B x + (2AB -A^3))(x^2 -A x + (A^2 - B)) = x^5 - x \pm 2759640 $$
Linear coefficient is $-1,$ so $A^2 B - B^2 - 2 A^2 B  + A^4 = -1,$ or $A^4 - A^2 B - B^2 = -1.$ 
Getting there; taking $x = A^2, y = B,$ we have $x^2 - xy - y^2 = -1,$ meaning that $x,y$ are consecutive Fibonacci numbers... as $(3,2), (8,5), (21,13), (55,34), (144, 89), \cdots$ If we allow $y=B$ negative instead, we get
$$ \color{purple}{(1,-2), (3,-5), (8,-13), (21,-34), (55, -89), (144, -233), \cdots}$$
Since $x$ needs to be a square, we will try $A^2 = 144,$ $A = \pm 12,$ $B =  -233$
$$ (x^3 + A x^2 + B x + (2AB -A^3))(x^2 -A x + (A^2 - B)) = x^5 - x \pm 2759640 $$
One selection is
$$ (x^3 - 12 x^2 - 233 x + 7320)(x^2 + 12 x + 377)  = x^5 - x + 2759640$$
Just negating $x$ gives
$$ (-x^3 - 12 x^2 + 233 x + 7320)(x^2 - 12 x + 377)  = -x^5 + x + 2759640,$$
$$ (x^3 + 12 x^2 - 233 x - 7320)(x^2 - 12 x + 377)  = x^5 - x - 2759640,$$
ALL OF THESE:
jagy@phobeusjunior:~$ ./mse
  a:  -12  b:  -233  2ab-a^3:  7320  a^2-b:  377  prod:  2759640
  a:  -12  b:  89  2ab-a^3:  -408  a^2-b:  55  prod:  -22440
  a:  -1  b:  -2  2ab-a^3:  5  a^2-b:  3  prod:  15
  a:  -1  b:  1  2ab-a^3:  -1  a^2-b:  0  prod:  0
  a:  0  b:  -1  2ab-a^3:  0  a^2-b:  1  prod:  0
  a:  0  b:  1  2ab-a^3:  0  a^2-b:  -1  prod:  0
  a:  1  b:  -2  2ab-a^3:  -5  a^2-b:  3  prod:  -15
  a:  1  b:  1  2ab-a^3:  1  a^2-b:  0  prod:  0
  a:  12  b:  -233  2ab-a^3:  -7320  a^2-b:  377  prod:  -2759640
  a:  12  b:  89  2ab-a^3:  408  a^2-b:  55  prod:  22440