Timeline for What are conditions such that the polynomial $x^2+1$ divides $p(y)+q(z)+ax+b=F(x,\, y, \,z)$?
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| Feb 20, 2021 at 15:56 | vote | accept | Safwane | ||
| Feb 20, 2021 at 12:16 | comment | added | Joe Silverman | I think your second approach is preferable, since it also works in characteristic 2. And indeed it will even work for $R[x,y,z]$ where $R$ is an arbitrary ring, since $x^2+1$ is monic, so the $x$ degree of $(x^2+1)G(x,y,z)$ is always at least 2 unless $G$ is identically 0. | |
| Feb 20, 2021 at 11:24 | history | answered | Per Alexandersson | CC BY-SA 4.0 |