Timeline for Sum of sets modulo a square
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 4, 2010 at 6:53 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
took out false claim
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| Aug 4, 2010 at 6:48 | comment | added | Aaron Meyerowitz | I suppose you are right. The approach I give shows (inter alia) that if A+B is a complete set of residues mod N=#A#B and both A and B contain 0 (no loss of generality in assuming that) then at least one of the two contains no elements relatively prime to N. | |
| Aug 4, 2010 at 5:50 | comment | added | Gerry Myerson | @Aaron, Fedor calls it an "easy lemma" so I infer he already has a proof and just wants a citation. | |
| Aug 4, 2010 at 4:25 | comment | added | Aaron Meyerowitz | Ok, this is messed up. {0,1,4,5,8,9}+{0,2,12,14,24,26} is a counter-example. But note that mod 6 the second set is {0,2,0,2,0,2}. I'll post a clearer answer and take this one down. | |
| Aug 4, 2010 at 3:26 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
Added easier proof
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| Aug 4, 2010 at 3:16 | history | answered | Aaron Meyerowitz | CC BY-SA 2.5 |