Given the quartic $$x^4−2 x^3+(^2−^2) x^2+2^2 x−^2 ^2=0$$ for integers $,$ and $$ where $^2=^2+^2$.
Can anybody prove that the Galois group cannot be $\mathbb{Z}_4$ when $a \neq b$? I have an elementary proof when $b = a$.
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Looking at the resolvent cubic may help. – Kenta S Sep 20 '22 at 13:52
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Please use LaTeX/MathJax formatting. – JonathanZ Sep 20 '22 at 13:56
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Have a look at the notes by K. Conrad on Galois groups of quartics. – Dietrich Burde Sep 20 '22 at 14:10
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I know a lot about Galois theory for quartics. But, this seems to a tricky one. – kent Sep 20 '22 at 14:33
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Where did this quartic come from? – Daniel Hast Sep 20 '22 at 15:10
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This quartic comes from the s-ocalled non-square ladder box problem known from recreational mathematics. – kent Sep 20 '22 at 16:16
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Even Newton, Descartes and Lagrange studied this quartic (at least for the b = a case)! – kent Sep 20 '22 at 17:39
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You mean the quartic in the https://en.m.wikipedia.org/wiki/Crossed_ladders_problem ? – Parcly Taxel Sep 23 '22 at 04:46
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No, this is the so-called crossed ladder problem. See https://www.tandfonline.com/doi/abs/10.1080/0025570X.1987.11977313 from MM 1987. I can send you the paper I wrote about the Galois group for the square ladder box problem (b= a case). – kent Sep 23 '22 at 05:39
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For references to the two types of ladder problems, see CMJ (The College Mathematics Journal) 2008, page 161. – kent Sep 24 '22 at 09:30