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In the case of cubic equations,

Casus irreducibilis occurs when none of the roots is rational and when all three roots are distinct and real (...) —Wikipedia's Casus irreducibilis article

So, $x^3-3x+1=0$ is definitely an example of casus irreducibilis.

Cardano's formula can express a rational root in terms of non-real radicals (yet it is unnecessary), as in this example: $x^3-15x-4=0$. Some (Working with casus irreducibilis) call this equation a casus irreducibilis, but this disagrees with the (supposed) Wikipedia definition (which is described below), as it has a rational solution, namely $x=\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$.

Does the question in the link just involve a misinterpretation of casus irreducibilis, or are there any trustworthy books or other sources which support the fact that equations like $x^3-15x-4=0$ (which yield a rational root through Cardano's formula, though unnecessarily, using roots of complex numbers) are casus irreducibilis?

I suppose that the Wikipedia definition should read

Casus irreducibilis occurs if and only if none of the roots is rational and if and only if all three roots are distinct and real (...)

instead, as this defines casus irreducibilis precisely.

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The "irreducibilis" part of casus irreducibilis is irreducibility over the rationals. Hence $x^3-15x-4=0$ is not casus irreducibilis.

The linked question, however, has not really misused the term.

My question is, using Cardano's method for casus irreducibilis...

implies that it is about a situation where the rational root test isn't used beforehand, and the equation is assumed casus irreducibilis; we want to tell if the root obtained from Cardano's formula is really a rational in disguise.

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  • $\begingroup$ In the linked paper, I found this sentence: "It might be added that there exist more complicated examples of the casus irreducibilis where there are no simple formulae for real roots of cubics, except by using imaginary numbers." just after solving $x^3-15x-4=0$, as if it suggested that $x^3-15x-4=0$ alone is a casus irreducibilis. Isn't that misleading? $\endgroup$
    – Wane
    Commented Apr 11, 2020 at 18:16
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    $\begingroup$ @Wane Yes, it's misleading. $\endgroup$ Commented Apr 11, 2020 at 18:17
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Historically casus irreducibilis simply meant the case where the discriminant is negative regardless of the existence of rational roots.

For instance, Lagrange's lectures on elementary mathematics discusses the irreducible case in connection with Bombelli's example $x^3 = 15x + 4$, which was chosen because it has an integer solution and so can be used to test the cubic formula.

The idea that "irreducible case" means the cubic polynomial has to be irreducible seems to be a modern idea from field and Galois theory, projected backward. Before Galois the concern was with generic solution formulas where the coefficients of the polynomial are parameters. Solvability and unsolvability of particular equations not in a parametric family was not really subject to a general theory until Galois placed it in the context of field extensions, Galois group actions and so on.

In short, the Wikipedia is probably wrong, but for the subsequent modern analysis of whether casus irreducibilis can be circumvented, of course the first thing to require is that there be no rational solution.

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  • $\begingroup$ Casus irreducibilis actually implies the existence of polynomails equations not solvable in radicals. For $$\cos(tan^{-1}(x))$$ is algebraic for any algebraic $x$. $\endgroup$ Commented Feb 25 at 21:55
  • $\begingroup$ If $r(\cos(x)+i\sin(x))$ is algebraic, $r\cos(x)$ and $r\sin(x)$ are algebraic. If $r\cos(x)$ and $r\sin(x)$ can not be expressed in radicals over the real numbers, neither can their minimal polynomials be solved by radicals! $\endgroup$ Commented Feb 25 at 22:01
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The real and imaginary parts of a cube root of an arbitrary complex number can not always be expressed as a finite sequence of arithmetic operations, square root extractions, and cube root extractions over the real numbers.

A proof is here.

https://en.wikipedia.org/wiki/Casus_irreducibilis

Implicit is that the real and imaginary parts of solutions to cubics and quartics can not always be expressed as a finite sequence of arithmetic operations and root extractions over the coefficients and constants.

This does not contradict the solvability of cubics and quartics. Each solution as a whole can be expressed as a finite sequence of arithmetic operations and root extractions over the coefficients and constants.

there is a pattern here.

For quadratics, the solutions' real and imaginary parts each can be expressed as a finite sequence of arithmetic operations and square root extractions over the coefficients and constants

For cubics and quartics, the solutions as a whole can be expressed as a finite sequence of arithmetic operations, square root extractions, and cube root extractions over the coefficients and constants, but their real and imaginary parts not necessarily so.

For quintics and higher, the solutions are not always expressible (even as a whole) as a finite sequence of arithmetic operations, nth root extractions, ($n$ being positive integers) over the coefficients and constants.

Another corollary.

Let $\arg(a,b)$ be the number of radians of an arc between ($a,b$) and ($\sqrt{a^2+b^2},0$)on a circle with center ($0,0$), and let $i^2=-1$

Then the cube roots of $a+ib$ can only be expressed in radicals if $\cos(\frac{\arg(a,b)}{3})$ is a solution to a polynomial belonging to a solvable group.

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Like a lot of terms in mathematics, the meaning of casus irreducibilis falls in the category of "it depends".

The historical importance of casus irreducibilis was that Cardano's method required the use of complex numbers even though all three solutions to the cubic were real. The existence of cubic equations of this form spurred the acceptance of complex numbers. Only after this were complex solution to quadratics accepted.

The confusion is because it was already known how to find rational roots of a cubic if any existed. So, an example like $x^3=15x+4$ would not have required Cardano's method, and thus would not have required the use of complex numbers. Thus, for the purpose of justifying the use of complex numbers, casus irreducibilis was the case where one was forced to use Cardano's method and where Cardano's method required complex numbers. Ergo, three distinct real roots, none of which is rational.

The example $x^3=15x+4$ allows one to illustrate the use of Cardano's method and how complex numbers are required in Cardano's method to solve an equation of this form. So, in this case, casus irreducibilis means a case where Cardano's method requires complex numbers, regardless of whether the equation could be solved without Cardano's method. Hence, three distinct real root (whether rational or not).

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  • $\begingroup$ Actually, the rational root theorem was discovered only after Cardano's formula. Long after. $\endgroup$ Commented May 14 at 15:59

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