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  "body": "\n## Statement\n\nLet $n \\ge 5$ be an integer.\n\n1. Let $t_1, \\dots, t_n$ be indeterminates and $K = \\Q(t_1, \\dots, t_n)$. The general polynomial $x^n + t_1 x^{n-1} + \\dots + t_{n-1} x + t_n \\in K[x]$ is not [[solvable by radicals|J7YS0DA3HQ]] over $K$.\n2. There is a polynomial of degree $n$ with rational coefficients that is not solvable by radicals over $\\Q$.\n\n## Remarks\n\nBy [[Galois' theorem on solvability|VHCWCFA3MX]], it suffices to find a Galois group that is not [[solvable|0P3KZK0FTV]].\nThe Galois group of the general polynomial of degree $n$ is the full [[symmetric group|0RJMNEN3BJ]] $S_n$, and [[$S_n$ is not solvable for $n \\ge 5$|0J53NE674D]].\nFor clause 2, the polynomial $x^5 - 4x + 2$ has Galois group $S_5$ over $\\Q$; see the [[quintic equation|S1WS18QEGJ]].\nMultiplying it by $x^{n-5}$ gives a polynomial of degree $n$ with the same nonsolvable roots.\nThe bound $n \\ge 5$ is necessary: every polynomial of degree at most $4$ is solvable by radicals, by the [[cubic formula|FB6J5WAT5V]] and the [[quartic formula|4CDFMKKZ01]].\n\n## Explanations\n\n### L1 human\n\nFor equations of degree $2$, $3$ and $4$ there are formulas that give the roots from the coefficients using arithmetic and roots.\nThe theorem says that no such formula can exist for degree $5$ or more.\nWorse, some single equations of degree $5$ with whole-number coefficients have roots that cannot be written with radicals at all.\n\n### L2 human\n\nThe theorem does not say that quintic equations have no solutions: by the fundamental theorem of algebra, each has five complex roots.\nIt says only that these roots are not always radical expressions in the coefficients.\nMany quintics, such as $x^5 - 2$, are still solvable by radicals; Galois theory decides which ones.\nThe roots can be expressed with other functions, for instance with the [[Bring radical|BSDCDJR6RQ]] or with elliptic modular functions.\n",
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