Talk:Square superparticular
I prefer a much bruter method to show the semiparticulars' superparticularity
If k = 4n: |
If k = 4n + 1: |
If k = 4n - 1: |
If k = 4n - 2: |
So we see for k = 4n, 4n + 1, and 4n - 2, a coefficient of 22 is canceled out, whereas for the k = 4n - 1, a coefficient of 2 is canceled out. FloraC (talk) 18:31, 7 February 2022 (UTC)
- I figured something like this would be possible but I sometimes get lost in how to simplify and group stuff in intermediate steps when I do it that way or I make mistakes simplifying/expanding so I tried to use the most intuitive approach I could think of. The observation that k=2n leads to a factor of 4 I think is a relatively intuitive explanation of why its superparticular for those cases. Also, it took me a little while but I believe S(k)/S(k+2) = (k+3)/(k-1) * k2/(k+2)2 is the equation you substituted into for the four cases? (Just arranged as one single fraction.) --Godtone (talk) 23:45, 7 February 2022 (UTC)
- Yup that's the equation I started from. We may go even more primitive if we want, such as from the definitions. It doesn't matter in the end. I consider the elementary algebraic operations an easy problem, i.e. the solution is guaranteed by a known routine. All the tricks on the other hand involve observation and intuition. I mean, it's totally reasonable for one to apply them to instantly gain the insight, but I take the liberty of assuming the readers' expectation here is a solid and sound presentation of the result. FloraC (talk) 00:55, 8 February 2022 (UTC)