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Feb 22, 2016 at 5:45 comment added Will Jagy @zy_ in that case, i would expect that enough effort would produce recipes for $5k + 2,3$ more work $7k+3,5,6.$ This is not the sort of problem where a finite number of such cases are going to finish the job. I did not see that Zagier really cared about all positive quaternaries of a given discriminant, just some compatible with the modular forms material.
Feb 22, 2016 at 5:16 comment added Y. Zhao Following Zagier, I can cover $p=8k-3$ with $$A = \left( \begin{array}{rrrr} 2 & 0 & 1 & 1 \\ 0 & 4 & 1 & -2 \\ 1 & 1 & 2k & 0 \\ 1 & -2 & 0 & 4k \end{array} \right)$$
Feb 22, 2016 at 2:55 vote accept Y. Zhao
Feb 22, 2016 at 3:25
Feb 21, 2016 at 21:51 history edited Will Jagy CC BY-SA 3.0
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Feb 21, 2016 at 21:34 history edited Will Jagy CC BY-SA 3.0
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Feb 21, 2016 at 21:22 history answered Will Jagy CC BY-SA 3.0