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I have read (probably) in Kanigel's book The Man Who Knew Infinity that S. Ramanujan devised his own method of solving the Quartic Equation after he learnt to solve the Cubic Equation. Does anyone know what exactly was Ramanujan's method of solving the Quartic Equation?


Update

It is now asked here at HSMSE.

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    $\begingroup$ This would be right at home at History of Science and Mathematics $\endgroup$ – user147263 Dec 5 '14 at 6:04
  • $\begingroup$ @WarmFuzzies: Thanks for the link. $\endgroup$ – user170039 Dec 5 '14 at 8:07
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    $\begingroup$ You must have it wrong. Compared to cubic equations, solving quadratic equations is child's play. $\endgroup$ – TonyK Dec 5 '14 at 11:02
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    $\begingroup$ The help center mentions "History and development of mathematics" as being on-topic for math.stackexchange. Also, here are some similar questions which were highly upvoted. I think this question is on-topic here. $\endgroup$ – littleO Dec 5 '14 at 11:13
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    $\begingroup$ This question has also been asked on History of Science and Math Stack Exchange, where it will likely find a better audience. $\endgroup$ – user642796 Dec 8 '14 at 22:18
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You can find it in Ramanujan's Notebooks IV by B. Berndt, Chap. 22 Elementary Results, Entry 20, p.31.

Ramanujan starts with this problem. Let $a,b,c,d$ be arbitrary. Solve the system,

$$x^2+ay = b\tag{20a}$$

$$y^2+cx = d\tag{20b}$$

Eliminating $y$, we find it is equivalent to,

$$a^2(d-cx) = (b-x^2)^2\tag{20.1}$$

Assume without loss of generality that $a=2$. Expanding this out,

$$x^4-2bx^2+4cx+(b^2-4d)=0\tag{20.1a}$$

By a simple linear substitution, the general quartic equation can be expressed in the depressed form,

$$x^4+px^2+qx+r = 0\tag{20.1b}$$

Equating coefficients of ${20.1a}$ and ${20.1b}$, you have a system of 3 equations in 3 unknowns {$b,c,d$}. Hence every quartic can be expressed in the form ${20.1}$. The problem then is to find $x$. Ramanujan defines,

$$x = \alpha+\beta+\gamma\tag{20.1c}$$

$$y = -(\alpha\beta+\beta\gamma+\gamma\alpha)$$

$$-c/2 = \alpha\beta\gamma$$

(If you are familiar with cubic equations, you'll already see where Ramanujan is going.)

Substitute $x,y,c$ into $(20a)$ and $(20b)$ keeping in mind $a=2$, then,

$$x^2+ay = \alpha^2+\beta^2+\gamma^2= b$$

$$y^2+cx = (\alpha\beta)^2+(\beta\gamma)^2+(\gamma\alpha)^2 = d$$

$$(-c/2)^2 = (\alpha\beta\gamma)^2$$

By elementary symmetric polynomials, we then conclude that $\alpha^2,\,\beta^2,\,\gamma^2$ are roots of the cubic equation,

$$t^3-bt^2+dt-c^2/4=0\tag{20.1d}$$

Of course, by solving $(20.1d)$, one can then find $\alpha,\,\beta,\,\gamma$. Using $(20.1c)$, we then further conclude that the four roots of the quartic are,

$$x = \alpha+\beta+\gamma, \quad\alpha-\beta-\gamma, \quad-\alpha-\beta+\gamma, \quad-\alpha+\beta-\gamma$$

P.S. This is similar to Euler's method where he solves a quartic as $x_i = \sqrt{y_1}\pm \sqrt{y_2}\pm \sqrt{y_3}$ and the $y_i$ are roots of a cubic. There's actually a generalization to this for solvable 8th deg eqns as $x_i = \sqrt{y_1}\pm \sqrt{y_2}\pm\dots\pm \sqrt{y_7}$ and the $y_i$ are roots of a 7th deg eqn. See this mathoverflow post.

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    $\begingroup$ @littleO: That similar question just happened to be mine. :) $\endgroup$ – Tito Piezas III Dec 5 '14 at 16:35
  • $\begingroup$ Can the same method be used to find a formula for the sextic? $\endgroup$ – Frank Oct 30 '16 at 18:29
  • $\begingroup$ @Frank: No, this method takes advantage of the fact that $n=4$ is a power of $2$. There is a similar approach for $n=8$ (see this MO post) though is solvable in radicals only if the octic has a solvable Galois group. $\endgroup$ – Tito Piezas III Oct 31 '16 at 1:33