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From what I've read, I think I have a symmetric quintic polynomial equation. I've spent months trying to solve it for one of it's variables $(x)$. It seems convertable to Bring-Jerrard form: $$2y^5x-2yx^5=z\implies x^5-y^4x+\frac{z}{2y}=\mathbf{x^5+ax+b=0}$$

I've bought and read books such as "Beyond the Quartic Equation" and now I'm reading "Classical Galois Theory, with examples" and "The Equation That Couldn't Be Solved: How Mathematical Genius Discovered the Language of Symmetry". I understand some of group theory but I don't know how to determine which group this equation falls into. I understand symmetry of geometric figures but I haven't made the connection about how symmetry helps solve equations.

I've visited dozens and dozens of sites about quintics but, like the books, most simply say what $is$ or $can$ be used to solve quintics and above. Finally, I found Example-4 in a "Russian Journal of Mathematical Research" article and it seems to hand me the answer (though I haven't tested it). The reported solutions are:

$$\\x_1=-(\frac{b}{16}+\sqrt{(((\frac{b}{16})^2-(\frac{b}{8}+((\frac{b}{8})^2-(\frac{b}{4}+\sqrt{(\frac{b}{4})^2-(\frac{b}{2}+\sqrt{(\frac{b}{2})^2-c})}})))))))$$

$$\\x_2=-(\frac{b}{16}-\sqrt{(((\frac{b}{16})^2-(\frac{b}{8}+((\frac{b}{8})^2-(\frac{b}{4}+\sqrt{(\frac{b}{4})^2-(\frac{b}{2}+\sqrt{(\frac{b}{2})^2-c})}})))))))$$

$$\\x_3=-(\frac{b}{16}+\sqrt{(((\frac{b}{16})^2-(\frac{b}{8}-((\frac{b}{8})^2-(\frac{b}{4}+\sqrt{(\frac{b}{4})^2-(\frac{b}{2}+\sqrt{(\frac{b}{2})^2-c})}})))))))$$

$$\\x_4=-(\frac{b}{16}-\sqrt{(((\frac{b}{16})^2-(\frac{b}{8}-((\frac{b}{8})^2-(\frac{b}{4}+\sqrt{(\frac{b}{4})^2-(\frac{b}{2}+\sqrt{(\frac{b}{2})^2-c})}})))))))$$

$$x_5=-\frac{c}{x_1x_2x_3x_4}$$

This format seems OK, resembling a trig solution for a cubic that I got help with here, and where the final formula to solve $mn^3-m^3n+D=0$ for $n$ became:

$$n_0=2\sqrt{\frac{m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\arccos{\biggl(-\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}$$ $$n_1=2\sqrt{\frac{m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\arccos{\biggl(\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}$$ $$n_2=n_1-n_0$$

This works in all cases I've tested; whichever produces an integer is what I'm looking for... and I appreciate the help I had understanding how it works.

Is the Russian formula correct? If so, can someone tell me how he got there in novice terms? I understood almost nothing between $x^5+ax+b=0$ and "$x_1=$".

If the Russian formula is not what I'm looking for, can someone help me find what group I'm dealing with, how symmetry matters, and how to proceed from there to a solution?

$\textbf{Update:}$ One comment said that anyone claiming to have a general formula to solve Bring-Jerrard form must be mistaken. I wonder if it would help to know that, in the equations: $2y^5x-2yx^5=z\implies x^5-y^4x+\frac{z}{2y}$ the "variable" $z$ will always be divisible by $60$ because it is the product of sides of a Pythagorean triple.

Here are the first $20$ expected triples of $(z,y,x)$ where $z$ and $y$ are input and $x$ is output. $$(60,2,1)\quad (480,3,1)\quad (780,3,2)\quad (2040,4,1)\quad (3840,4,2)\quad (4200,4,3)\quad (6240,5,1)\quad (12180,5,2)\quad (14760,5,4)\quad (15540,6,1)\quad (16320,5,3)\quad (30720,6,2)\quad (33600,7,1)\quad (40260,6,5)\quad (43740,6,3)\quad (49920,6,4)\quad (65520,8,1)\quad (66780,7,2)\quad $$

So, how would one go about solving $2y^5x-2yx^5=60\quad$ or $\quad 2y^5x-2yx^5=780$? Perhaps it would indicate an approach to solving this particular quintic equation.

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  • $\begingroup$ Where's the equation you want to solve? The equations above contain multiple variables. $\endgroup$ – Allawonder Dec 31 '19 at 20:45
  • $\begingroup$ @poetasis Do you mean $2x^5y-2y^5x+z=0$? What you wrote in the comment is just $z=0$... $\endgroup$ – YiFan Jan 1 at 10:35
  • $\begingroup$ This is not a symmetric polynomial. $\endgroup$ – Robin Carlier Jan 1 at 10:57
  • $\begingroup$ @YiFan Thanks for the correction. My fingers were fat last night. $\endgroup$ – poetasis Jan 1 at 11:05
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    $\begingroup$ I carefully inputed the solution you gave in a symbolic calculator, and it does not reduces to 0. In fact, I highly doubt that you will ever find a working solution by radical of the Bring-Jerrard normal form. Any quintic can be brought to such a form, and a solution of a general quintic of this form would yield a solution of a general quintic, which is known to be impossible by Abel-Ruffini's theorem. All the articles claiming to have a solution must have mistakes at some points. $\endgroup$ – Robin Carlier Jan 1 at 11:23

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