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Hi. I chose a simple equation to work with: $x^5+x^4+x^3+x^2+x+1=0$ (which is solvable and $x=-1$).

the amounts of $p$, $q$, $r$ and $s$ where $\frac{3}{5}$, $\frac{14}{25}$, $\frac{87}{125}$ and $\frac{2604}{3125}$ respectively.

I calculated $P$ to be $P=z^3-15z^2-69z+1235$ and $\Delta$ to be $\Delta=1296$.

as the picture says, $P^2-1024z\Delta$ (Cayley's resolvent) should have rational root(s) in $z$ in order to the first qintic function be solvable.

but the problem is that $P^2-1024z\Delta=0$ is $z^6-30z^5+87z^4+4540z^3-32289z^2-1497534z+1525225=0$ which is even worse and impossible to solve!

what should I do now?!

  • I know its factorization friend. I just want to know how to use the Cayley's resolvent for this equation or other quintics. @DietrichBurde –  Feb 12 '20 at 14:16
  • If it was unnesessary, Cayley wouldn't introduce it. @DietrichBurde –  Feb 12 '20 at 14:18

2 Answers2

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Hint:

We have the following factorization: $$ z^6-30z^5+87z^4+4540z^3-32289z^2-1497534z+1525225= (z^2 + 22z + 169)(z^2 - 26z + 361)(z - 1)(z - 25). $$ Of course, the splitting field of $$ x^5+x^4+x^3+x^2+x+1=\frac{x^6-1}{x-1} $$ is $\Bbb Q(\zeta_6)$, so that the Galois group is solvable.

Dietrich Burde
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  • thank you. please tell me how did you factorize it? –  Feb 12 '20 at 14:27
  • You know this already ("I know its factorization friend"), or not? I used the rational root test to find the roots at $1$ and $25$ first and then wrote the product of two quadratic polynomials for the quartic. – Dietrich Burde Feb 12 '20 at 14:28
  • "I know its factorization friend" was for $x^5+x^4+x^3+x^2+x+1$. NOT for $z^6-30z^5+87z^4+4540z^3-32289z^2-1497534z+1525225$ –  Feb 12 '20 at 14:31
  • But you can use both factorizations for this example and you only need to factorize $x^5+\cdots +1$ here. For other examples, it makes sense to use Cayley's resolvent. – Dietrich Burde Feb 12 '20 at 14:33
  • how did you conclude you should find the roots at 1 and 25? and not for example in 1 and 24? –  Feb 12 '20 at 14:34
  • By the rational root test, we should look for divisors of $1525225=5^2\cdot 13^2\cdot 19^2$, so $24$ is impossible. – Dietrich Burde Feb 12 '20 at 14:34
  • can Cayley's resolvent always tell us the root of any given solvable quintics? or just tells us it is solvable or not? –  Feb 12 '20 at 14:37
  • Don't forget, if the quintic is not solvable, then it means that we cannot find the roots (by radicals). – Dietrich Burde Feb 12 '20 at 14:38
  • my last question. you said we should look for divisors for rational root test. but $1525225$ has $27$ divisors. why do we just choose $1$ and $25$? or why don't we choose $13^2$? @Dietrich Burde –  Feb 12 '20 at 14:44
  • One way to cut down on trials is to observe that since $1024z\Delta$ is equated to a squared quantity, a proposed rational root for $z$ must give a square product with $\Delta$, which constrains sign as well as magnitude. Here $z$ would have to be positive and, with $\Delta$ a perfect square, so is $z$. Then the only surviving candidates below $100$ are $1,25$. Candidates larger than $100$ may be certified to fail by $z^6$ absolutely dominating the other terms. – Oscar Lanzi Jul 19 '23 at 12:35
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First off, Cayley's resolvent is meant for irreducible quintics, which the given example is not. If the quintic is reducible, then it can always be solved by radicals through the solution of the lower-degree factors and yet Cayley's resolvent may fail to give a rational root. For instance, if we blindly apply Cayley's resolvent to the equation

$x^5-x^3+x-1=0,$

we ultimately fail to find a rational root to the resolvent; yet this quintic is solvable by factoring it as $(x-1)(x^4+x^3+1)$ and applying the usual radical-based algorithm to the quartic factor.

When applying Cayley's resolvent to an irreducible quintic equation, you do not need to fully solve the resolvent sextic. You need only to check for a rational root. If none exists, you're done with the method; the quintic is not solvable. If the rational root does exist, then the quintic will be solvable and you're good if that is all you need to know. However, going on to actually get the quintic roots requires putting the resolvent root and the depressed coefficients into a quite complicated formula.

It is good practice to find the constant term in the cubic polynomial $P$ first, because this could be zero. If so, then the full sextic resolvent also has a zero constant term and we can immediately identify zero as the required rational root (also the formula for the quintic roots themselves becomes simplified).

See this answer for an application to the depressed quintic equation $x^5+20x^3+20x^2+30x+10=0$.

Oscar Lanzi
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