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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.

poetasis
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  • Where's the equation you want to solve? The equations above contain multiple variables. – Allawonder Dec 31 '19 at 20:45
  • @poetasis Do you mean $2x^5y-2y^5x+z=0$? What you wrote in the comment is just $z=0$... – YiFan Tey Jan 01 '20 at 10:35
  • This is not a symmetric polynomial. –  Jan 01 '20 at 10:57
  • @YiFan Thanks for the correction. My fingers were fat last night. – poetasis Jan 01 '20 at 11:05
  • @Robin Carlier Do you mean the "wrong" equation in my now-deleted comment or the "correct" equation in YiFan's comment? – poetasis Jan 01 '20 at 11:08
  • Both. A symmetric polynomial P(x, y, z) is a polynomial of several variable that remains invariants when permuting the factors. In your polynomial, if you change $x$ for $z$, the polynomial is not the same, same if you switch $x$ and $y$. If you were talking about the polynomial seen as a polynomial of the single variable $x$, then it is true that it is symmetric, but every singlevariate polynomial is, so it makes no difference. –  Jan 01 '20 at 11:11
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    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. –  Jan 01 '20 at 11:23
  • @poetasis are you looking for integer solutions? It is unlikely that the quintic can be solved even with the extra conditions on the coefficients. If you are looking for integer solutions, given $z$, we know $x,y$ must divide $z$ and that leaves finitely many cases to consider. For example $2y^5x-2xy^5=30$ has no integer solutions since the left hand side has a factor $x^2+y^2$ and the only factors of 15 that are a sum of two squares are 1 and 5. Since $x,y\neq 0$, that would mean $y=2$ and $x=1$ which doesn't solve the equation. Maybe arguments like this be more useful in this setting. – Marco Jan 02 '20 at 20:16
  • @ Marco You might have the right idea but can we try it again with $z=60k$, where $k$ is an integer multiplier? The factors on the left side are $(2xy)(x-y)(x+y)(x^2+y^2)$ but how does that help us when we know only the value of $z$ and a range of values of $y$ to try to see if any yield an integer $x$? – poetasis Jan 02 '20 at 21:48

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