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I would like to know if there is a close-form formula for $x^5 + x = y$.

For $x^2 + x = y$, we have:

$$ \begin{aligned} &x = \frac{-1 + \sqrt{1+4y}}{2} \end{aligned} $$

For $x^3 + x = y$ we have: $$ \begin{aligned} &x = \frac{1}{2}\sqrt[3]{2}^2\left(\sqrt[3]{\frac{\sqrt{81y^2+12}}{9}+y}+\sqrt[3]{\frac{-\sqrt{81y^2+12}}{9}+y}\right) \end{aligned} $$

For $x^4 + x = y$ we have: $$ \begin{aligned} &\gamma = \sqrt[3]{\frac{1}{2} + \sqrt{\frac{1}{4}+\frac{(4y)^3}{27}}} + \sqrt[3]{\frac{1}{2} - \sqrt{\frac{1}{4}+\frac{(4y)^3}{27}}}\newline &x=\frac{1}{2}\left(-\sqrt{\gamma} + \sqrt{\gamma - 2\left(\gamma - \frac{1}{\sqrt{\gamma}}\right)}\right) \end{aligned} $$

Henry
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isedgar
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    In short, there is no general solution for polynomials with degree $>4$ in terms of radicals. Some quintics may be solved using other identities, but there does not exist any general closed form. For more you can read up the Abel-Ruffini theorem. – Sahaj Dec 05 '23 at 17:19
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    You mean a closed formula for $x$ in terms of $y,$ presumably. You've already written a closed formula. – Thomas Andrews Dec 05 '23 at 17:27
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    Galois group shows that $x^5+x=3$ cannot be solved in radicals. – GEdgar Dec 05 '23 at 17:31
  • Why solve for $x$ and not for $y$ then? Instead of, say, $y=3$ you take $y$ as a variable. So the closed formula is $y=f(x)=x^5+x$. You could also consider $x^5+x=y^3$, and solve this over the integers. Actually, are $x$ and $y$ integers? You say nothing about the domain. – Dietrich Burde Dec 05 '23 at 19:17
  • Hasn’t this question been asked a few times before on this site? – Тyma Gaidash Dec 05 '23 at 19:30

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Not in terms of radicals, at least: We can express all of the roots of a polynomial $p$ in terms of radicals if and only if the Galois group $\operatorname{Gal}(p)$ is solvable. But, for example, a solution of $x^5 + x = 3$ is a root of $x^5 + x - 3$, and $\operatorname{Gal}(x^5 + x - 3) \cong S_5$, which is not solvable. A fortiori there is no formula for general $y$ that expresses the solutions $x$ of $x^5 + x = y$ in terms of radicals. By contrast the Galois group of any polynomial of degree $\leq 4$ is solvable hence admits such an expression for its roots.

The function $x^5 + x$ is a strictly increasing function of $x$ with image all real numbers, so there is a unique value $x$ such that $x^5 + x = y$, hence that value is a function of $y$. This function is essentially the Bring radical, which defines $\operatorname{BR}(a)$ to be the unique real root of $x^5 + x + a$, and in terms of this function, the equation $x^5 + x = y$ has unique solution $\beta := \operatorname{BR}(-y)$.

Long division gives $$x^5 + x - y = (x - \beta) [x^4 + \beta x^3 + \beta^2 x^2 + \beta^3 x + (\beta^4 + 1)] ,$$ and in principle you can now apply methods for solving quartic polynomials to the second factor to express the roots of $x^5 + x - y$ in terms of $\beta$ and radicals.

Travis Willse
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