3

I read on Wikipedia that x^5 -x -1 = 0 has a real root, but that you can't express it in radicals. So I thought maybe all of the x^5 -x -A =0 don't have a real root that can be expressed as a radical and I couldn't find anything online about it. So I saw that the answer of the -1 case was an irrational number and my plan was to force it into another irrational number that I did know that was really close. So I kept evaluating different A values to get me closer to root 2. Eventually I got to x^5 -x -4.24264068711 = 0 which gives a real root very close to root 2. So then I figured out that the closed form of 4.24264068711 is actually 3(2)^.5 or the square root of 18.

My question is why does x^5 -x -(18)^.5 = 0 have a nice real root that you can express with radicals, while x^5 -x -1 = 0 does not have a nice radical form like that?

Neil
  • 880
  • Just to confirm your result, note that $\sqrt{2}^5 = (\sqrt 2)^4 \sqrt 2 = 4 \sqrt 2$, so putting $x = \sqrt 2$ into $x^5 - x - 3\sqrt 2$ does indeed get you zero. – Ben Millwood Apr 06 '15 at 23:39
  • Solvability by radicals of a polynomial is related to the solvability of the Galois group. Galois proved that an irreducible polynomial of prime degree over a field, is solvable by radicals if and only if its Galois group is contained in the affine linear group on p elements, equivalently, if any two of its roots generate its splitting field. You can look for the Galois group of this polynomial (I think $x^{5}-x-1$ is irreducible oer $\mathbf{Q}$?) – mich95 Apr 06 '15 at 23:39

1 Answers1

2

You shouldn't find it surprising that you were able to pick an $A$ value to get a particular root you wanted. Here's a more general recipe: suppose you want $c$ to be a root of $x^5 - x - A$, then you can just choose $A$ to be $c^5 - c$, then of course $x^5 - x - (c^5 - c)$ is going to be zero when evaluated at $x = c$. You can do a similar trick if you control any coefficient of any polynomial (unless you want $0$ to be a root, in which case the coefficient you control needs to be the constant coefficient).

This should also make it clear that some $A$ values will be the "right" $A$ values for scary (non-radical) roots, whereas some will be right for nice (radical) roots.

Ben Millwood
  • 14,211
  • If 1 = A = c^5 - c, then why can't you divide out the c and get c(c^4-1) and then use the quartic formula to get the nice radical value? – Neil Apr 06 '15 at 23:56
  • 1
    @Neil Where would you use the quartic there? Dividing by $c$ just gives you $c^4-1=\frac1c$, and that's not a quartic equation... – Steven Stadnicki Apr 06 '15 at 23:58
  • Yeah that's right Steven, my bad. I was thinking that you could do c^4-1 =1 instead and ignore the outside c. – Neil Apr 06 '15 at 23:59
  • Is the reason it doesn't work because there is some other special math operation being used? For example: roots are a kind of exponent and exponents are a kind of multiplication and multiplication is a type of addition. So all the nice numbers use some kind of addition, but the ones that aren't clean have some piece in them that isn't using addition? – Neil Apr 07 '15 at 00:07
  • Not exactly. Multiplication is strongly related to addition, but they're not the same thing. The truth is there aren't "nice numbers" and "scary numbers", but just numbers that are nice for some particular purpose, or numbers that are hard to deal with using some particular tools. What that purpose is or those tools are is a consequence of what you happen to be doing at the time. One common set of tools is the set you outlined, but it wouldn't be totally unreasonable to include the tool "finding roots of arbitrary polynomials", in which case these numbers are "nice" too. – Ben Millwood Apr 11 '15 at 23:46