Timeline for long enough interval of integers to solve a simultaneous congruence
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 8, 2013 at 1:06 | comment | added | Noam D. Elkies | P.S. About the Combinatorial Nullstellensatz: fearsome or not, it would not be not fair game for the Putnam exam. | |
| Mar 7, 2013 at 15:32 | comment | added | Noam D. Elkies | @Günter Rote: I need the $m_j$ to be distinct mod $a$ because they enter the final argument as exponents in $w^{m_j}$ and I need those roots of unity to be distinct. To see that this is necessary, try to apply the argument when some $A_i = \emptyset$, for which the result is false but everything else in the proof works. | |
| Mar 6, 2013 at 18:21 | comment | added | Günter Rote | @Noam. Clarification: I was not puzzled because I did not see that the product has "distinct" exponents, after expanding the product and before collecting terms with the same exponent. But why is this fact needed? | |
| Mar 6, 2013 at 5:56 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
Spell out why the $m_j mod a$ are distinct
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| Mar 5, 2013 at 23:11 | comment | added | Labrador | So it all boils down to the nonzeroness of the Vandermonde determinant... like many other things in this world :). Thanks a lot, I have to say this is a sad day for me. I'd thought the correct solution might have about this flavor and length (with polynomials and roots of unity) but I didn't have the faith (or brains) to push it through. Nicely done. For those who are curious, this problem stems from the conjecture put forth in the paper "On the entry sum of cyclotomic arrays" by myself and Coppersmith. It doesn't solve that conjecture on its own, but who knows. | |
| Mar 5, 2013 at 22:55 | vote | accept | Labrador | ||
| Mar 5, 2013 at 22:55 | history | bounty ended | Labrador | ||
| Mar 5, 2013 at 20:42 | comment | added | Günter Rote | Nice! I am puzzled by the remark that the $N$ monomials have distinct exponents. If they were not distinct, just collect terms and let them cancel if they want, and let $m_j$ only refer to the nonzero terms. This would reduce the number of terms of $P(X)$, and the argument goes through (with a reduced number $N$) as long as the polynomial does not become identically zero. But this is ok, as we know that the constant term is nonzero because it has abs. value 1. (And anyway, who is afraid of the Combinatorial Nullstellensatz? It doesn't even need such advanced stuff as "complex numbers;-) | |
| Mar 5, 2013 at 20:11 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
Add Putnam comment and correct typo in definition of A
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| Mar 5, 2013 at 17:04 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |