I recently saw a very hideous closed form for a quartic equation here: Does a closed form solution exist for $x$?

For fun, I'm wondering about surprisingly ugly solutions/ complicated machinery needed to problems that are simply stated.

Clearly, they all don't have to be algebra-based.

I'm trying to get a sense for how particular different mathematic methods are necessary, and interesting. Even for easily stated problems.


closed as primarily opinion-based by user147263, user296602, colormegone, MJD, Stefan Mesken Mar 16 '16 at 4:56

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  • $\begingroup$ It depends what you mean by an ugly solution. I could ask you to solve the equation $2x = 2$ and you could give me the answer "Clearly, $x = \frac{\Gamma\left(\frac{1}{2}\right)}{\sqrt\pi}$." Would that be an ugly solution, or are you looking for surprisingly "over the top" solutions to otherwise simple problems? $\endgroup$ – ÍgjøgnumMeg Mar 16 '16 at 1:58
  • 2
    $\begingroup$ Nice response to the irritant bot. $\endgroup$ – Rob Arthan Mar 16 '16 at 2:01
  • $\begingroup$ @ed_4434 more like the latter! Have something in mind? $\endgroup$ – Andres Mejia Mar 16 '16 at 2:07
  • $\begingroup$ If you are looking for surprisingly over the top solutions to otherwise easily stated problems, look at Fermats last theorem, possibly THE defining example of this kind of problem! Easy to state but it took about 300 years to prove $\endgroup$ – ÍgjøgnumMeg Mar 16 '16 at 9:09
  • $\begingroup$ Agreed. I actually find most open problems in number theory to be of this nature, part of what makes it a fascinating subject $\endgroup$ – Andres Mejia Mar 16 '16 at 13:29

Classify pairs of commuting matrices up to a change of basis.

This sounds like an extension of Jordan normal form, since the matrices should be simultaneously Jordan-izable or something.

In fact it is the canonical example of a "wild" linear algebra problem that is as hard as classifying $k$-tuples of (noncommuting) matrices for all $k$. No reasonable solution is expected to exist.


Given a polynomial, $p(x)$, with integer coefficients, what numbers, $z$, satisfy the equation $p(z)=0$.

In the case of small degree polynomials, we have the ability to write down the solutions to polynomial equations with integer coefficients using integers, the symbols $(,),+,-,\cdot,/,$ and ^, where the last indicates exponentiation. In degree $2$, this gives rise to the well known equation $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

In degree 5 or greater, this is suddenly no longer true. There are fifth-degree polynomials with solutions that cannot be exactly described in any "reasonable" way.

  • $\begingroup$ I agree! I was extremely surprised when i first learned of this $\endgroup$ – Andres Mejia Mar 16 '16 at 2:06

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