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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?

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  • $\begingroup$ It also factors when $a=0$, $a=\pm 30$, $a=\pm 240$, $a=\pm 1020$, and so on - the cases with an integer root. There must be something more to the statement excluding those cases. $\endgroup$
    – jmerry
    Feb 27, 2019 at 3:11
  • $\begingroup$ You're right. "[The form] has solutions in radicals if and only if it has an integer solution or r is one of ±15, ±22440, or ±2759640, in which cases the polynomial is reducible." $\endgroup$
    – KKZiomek
    Feb 27, 2019 at 3:13
  • $\begingroup$ I would say that factorization into radicals is a special case of factorization, and the fact that there is no factorization of some polynomial into radicals does not imply there is no factorization at all. For any real number $a,$ the equation $x^5-x+a=0$ has at least one real root; if $\beta$ is the least root, then $x^5-x+a=(x-\beta)(x^4+\beta x^3+\beta^2x^2+\beta^3x+(\beta^4-1)).$ That's a factorization. The catch is that $\beta$ cannot be written in radicals except in the special cases named in the question or when $a=n^5-n$ for some integer $n.$ $\endgroup$
    – David K
    Feb 27, 2019 at 3:50
  • $\begingroup$ The finiteness of the list and the method of factoring come down to the fact that the largest square Fibonacci number is $144$ $\endgroup$
    – Will Jagy
    Feb 27, 2019 at 15:50
  • $\begingroup$ I would also consider doing modular factorization first? Factorization w.r.t. a prime modulus is reasonably fast, and we can then Chinese Remainder Theorem -combine several primes. Also Hensel lifting a factorization to a prime power modulus should quickly cull the search space. Here it is easy to see that modulo two the factorization is $(x^2+x+1)(x^3+x^2+1$, so we already know that a quadratic times a cubic is the only alternative. Need to try this some other day to see how quickly it is resolved. $\endgroup$ Feb 27, 2019 at 22:22

4 Answers 4

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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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The statement on Wikipedia cites a 1998 paper dealing with the solutions of $x^k-x=n$ for nonnegative integer $n$ and $k=3,4,5$. In turn this paper cites another paper by Rabinowitz ("The Factorization of $x^5\pm x+n$", Math. Magazine 61.3 (1988), 191–193) showing that if $x^5-x-r$ has no linear factor yet is reducible over the integers, then one of the two equations below is true: $$r=\pm F_{2j-1}F_{2j}\sqrt{F_{2j+2}}\tag1$$ $$r=\pm F_{2j}F_{2j+1}\sqrt{F_{2j-2}}\tag2$$ The three specific $r$ that cause the quintic to split (into a quadratic and cubic factor) then correspond to the only nonzero square Fibonacci numbers of even index, $F_2=1$ and $F_{12}=144$:

  • $(2)$ with $j=2$ gives $r=\pm15$
  • $(1)$ with $j=5$ gives $r=\pm22440$
  • $(2)$ with $j=7$ gives $r=\pm2759640$

Because there are only three special quintics, memorising the quadratic factors is feasible:

  • $x^2\pm x+3$ divides $x^5-x\pm15$
  • $x^2\mp12x+55$ divides $x^5-x\pm22440$
  • $x^2\pm12x+377$ divides $x^5-x\pm2759640$

Then long division should suffice to find the cubic cofactor.

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I see there is an answer to 15. I tried, it appears there is no integer root to $x^5 - x \pm 22440.$ As 7 gives 16800 but 8 gives 32760. We arrive at $$ (x^3 + A x^2 + B x + C)(x^2 + D x + E) = x^5 - x \pm 22440 $$ 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 $A+D = 0$ $$ (x^3 + A x^2 + B x + C)(x^2 -A x + E) = x^5 - x \pm 22440 $$ Next the cubed term is zero, so $E-A^2 + B = 0, $ or $E = A^2 - B.$ $$ (x^3 + A x^2 + B x + C)(x^2 -A x + (A^2 - B)) = x^5 - x \pm 22440 $$ Next $x^2$ has 0, or $A^3 - AB -AB + C = 0,$ or $C = 2AB - A^3,$ $$ (x^3 + A x^2 + B x + (2AB -A^3))(x^2 -A x + (A^2 - B)) = x^5 - x \pm 22440 $$ 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 $(3,2), (8,5), (21,13), (55,34), (144, 89).$ Since $x$ needs to be a square, we will try $A^2 = 144,$ $A = 12,$ $B = \pm 89$

One selection is $$ (x^3 + 12 x^2 + 89 x + 408)(x^2 - 12 x + 55) = x^5 - x + 22440$$ Just negating $x$ gives $$ (-x^3 + 12 x^2 - 89 x + 408)(x^2 + 12 x + 55) = -x^5 + x + 22440,$$ $$ (x^3- 12 x^2 + 89 x - 408)(x^2 + 12 x + 55) = x^5 - x - 22440,$$

So there you go. We were able to use the first Fibonacci number that is also a square. Bigger than $1$ I guess.

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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
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