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I wish to show if for sufficiently large integers N, and a given prime p, the polynomial $x(x − N p^2)(x + N p^2)(x^2 + N^2p^4) + p $ can be solved by radicals. I think the answer is no but I am not sure how to prove~

Homaniac
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  • What do you mean by "solve"? Is it an equation? Did you leave out an $=$ sign somewhere? – bof Nov 09 '18 at 06:32
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    The quintic is irreducible by Eisenstein. So if it has three real roots then we are done. Easy check. – user10354138 Nov 09 '18 at 06:40
  • Ah I see, what are the three real roots tho – Homaniac Nov 09 '18 at 07:07
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    You can easily show it has at least three real roots by intermediate value theorem. To see why it cannot have five real roots, compute the discriminant. – pisco Nov 09 '18 at 11:59
  • What user10354138 and pisco said. It should be clear that for very large $N$ that last term $+p$ will only move the zeros by a tiny amount (the obvious of roots of the rest of it are simple, and the derivative at those roots is quite large => the zeros won't move much. – Jyrki Lahtonen Nov 11 '18 at 06:27
  • Not unlike Cam McLeman's fine solution here. – Jyrki Lahtonen Nov 11 '18 at 06:32
  • So let $L$ be the splitting field of that polynomial. Let $G = \operatorname{Gal}(L/Q)$. Then $5 \mid |G|$ because the polynomial is irreducible. So $G$ contains a 5 cycle. Next we show that $G$ contains a transposition too. (I'm guessing it is the restriction of complex conjugation to L but I am not sure.) Then these 2 elements generate $S_5$, so $G = S_5$. Then $S_5$ not solvable implies $L$ is not solvable. Is this right? – eatfood Nov 10 '19 at 18:31

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