Talk:Square superparticular

I prefer a much bruter method to show the semiparticulars' superparticularity

Demonstration of semiparticulars' superparticularity

If k = 4n: [math] \begin {align} {\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n)/{\rm S}(4n + 2) \\ &= \frac {((4n) + 3)(4n)^2}{((4n) - 1)((4n) + 2)^2 } \\ &= \frac {(4n + 3)(4n)^2}{(4n - 1)(4n + 2)^2} \\ &= \frac {(4n + 3)(2n)^2}{(4n - 1)(2n + 1)^2} \\ &= \frac {16n^3 + 12n^2}{16n^3 + 12n^2 - 1} \end {align} [/math]	If k = 4n + 1: [math] \begin {align} {\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n + 1)/{\rm S}(4n + 3) \\ &= \frac {((4n + 1) + 3)(4n + 1)^2}{((4n + 1) - 1)((4n + 1) + 2)^2 } \\ &= \frac {(4n + 4)(4n + 1)^2}{4n(4n + 3)^2} \\ &= \frac {(n + 1)(4n + 1)^2}{n(4n + 3)^2} \\ &= \frac {16n^3 + 24n^2 + 9n + 1}{16n^3 + 24n^2 + 9n} \end {align} [/math]
If k = 4n - 1: [math] \begin {align} {\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n - 1)/{\rm S}(4n + 1) \\ &= \frac {((4n - 1) + 3)(4n - 1)^2}{((4n - 1) - 1)((4n - 1) + 2)^2 } \\ &= \frac {(4n + 2)(4n - 1)^2}{(4n - 2)(4n + 1)^2} \\ &= \frac {(2n + 1)(4n - 1)^2}{(2n - 1)(4n + 1)^2} \\ &= \frac {32n^3 - 6n + 1}{32n^3 - 6n - 1} \end {align} [/math]	If k = 4n - 2: [math] \begin {align} {\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n - 2)/{\rm S}(4n) \\ &= \frac {((4n - 2) + 3)(4n - 2)^2}{((4n - 2) - 1)((4n - 2) + 2)^2 } \\ &= \frac {(4n + 1)(4n - 2)^2}{(4n - 3)(4n)^2} \\ &= \frac {(4n + 1)(2n - 1)^2}{(4n - 3)(2n)^2} \\ &= \frac {16n^3 - 12n^2 + 1}{16n^3 - 12n^2} \end {align} [/math]

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)