For a quadratic polynomial there exists a formula for its roots. I read that similarly for polynomials of degree 3 and 4 there also exists such a formula but that no such formulas exist for polynomials of degree 5 or higher.

Does there exist a proof that no such formulas for $\ge 5$ can exist or is it that they have not been found yet?

  • $\begingroup$ Yes, such a proof exists, as answered by Jared. It was proven by the Norwegian mathematician Niels Henrik Abel (1802-1829). $\endgroup$ Aug 12, 2013 at 8:13
  • $\begingroup$ Actually, formulas do exist for some- but-not-all polynomials of degree 5-or-higher. This depends on the Galois group of the polynomial. This is part of Galois Theory; you calculate the Galois group and decide if it is solvable, I believe. $\endgroup$
    – FBD
    Aug 12, 2013 at 8:13
  • $\begingroup$ Am I correct in thinking that, even though no general formula exists (i.e. for all possible solutions), this does not mean to say that the solutions themselves do not exist? $\endgroup$
    – pshmath0
    Aug 12, 2013 at 8:16
  • 2
    $\begingroup$ @pbs, all this means there is no general solution by means of radicals (say, as in the quadratic, cubic or quartic case) of a quintic equation. $\endgroup$
    – DonAntonio
    Aug 12, 2013 at 8:43
  • $\begingroup$ See the book Abel's Proof by Peter Pesic. $\endgroup$
    – lhf
    Aug 15, 2013 at 19:24

2 Answers 2


There is a proof, and it is quite beautiful. The result is known as the Abel-Ruffini Theorem, and the standard proof of this fact uses Galois theory and the fact that the alternating group on $n\ge 5$ symbols is not solvable. This wasn't the original proof, but it is the most elegant. The link provided contains a brief outline of the proof based on Galois's ideas.


There does exists a proof that polynomials with degree greater than 4 cannot be solved with radicals, the proof relies heavily on Galois theory. See for example here.


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