Timeline for Sum of sets modulo a square
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2010 at 7:25 | comment | added | Aaron Meyerowitz | I show that if $x^{n^2}-1$ does divide A(x)B(x) and A(x) is equally distributed mod n then B(x) is not equally distributed mod n. | |
| Aug 4, 2010 at 7:24 | vote | accept | Fedor Petrov | ||
| Aug 4, 2010 at 7:20 | comment | added | Fedor Petrov | Oh, Aaaron, thanks! That's what I actually needed. Gjergji, I think Aaaron is correct: we assume that $A+B$ is full set of residues modulo $N$ ($N=n^2$ in my case), then we have such equality and that's it. | |
| Aug 4, 2010 at 7:14 | history | edited | Gerry Myerson | CC BY-SA 2.5 |
improved formatting
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| Aug 4, 2010 at 7:03 | comment | added | Gjergji Zaimi | I don't see how any of the facts you wrote help with this problem. Here we need to show that $x^{n^2}-1$ does not divide $A(x)B(x)$ in your notation... | |
| Aug 4, 2010 at 6:44 | history | answered | Aaron Meyerowitz | CC BY-SA 2.5 |