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For ultraparticulars, we want to show that S''k''/S(''k'' + 1) = ((''k'' + 2)/(''k'' - 1)) / ((''k'' + 1)/''k'')<sup>3</sup>:
For ultraparticulars, we want to show that S''k''/S(''k'' + 1) = ((''k'' + 2)/(''k'' - 1)) / ((''k'' + 1)/''k'')<sup>3</sup>:


S''k''/S(''k'' + 1) = (''k''<sup>2</sup>/(k<sup>2</sup> - 1))/((''k'' + 1)<sup>2</sup>/((''k'' + 1)<sup>2</sup> - 1))
<math>
\begin {align}
{\rm S}k/{\rm S}(k + 1) &= \frac {k^2/(k^2 - 1)}{(k + 1)^2 / ((k + 1)^2 - 1)} \\
&= \frac {k^2}{(k + 1)(k - 1)} \cdot \frac{k^2 + 2k}{(k + 1)^2} \\
&= \frac {k^2}{(k + 1)^3 (k - 1)} \cdot (k + 2)k \\
&= \frac {k^3}{(k + 1)^3} \cdot \frac{(k + 2)}{(k - 1)} \\
&= \frac {(k + 2)/(k - 1)}{((k + 1)k)^3}
\end {align}
</math>


: = k<sup>2</sup>/(k+1)/(k-1) * (k<sup>2</sup> + 2k)/(k+1)<sup>2</sup>
For semiparticulars, we want to show that S''k''/S(''k'' + 2) = ((''k'' + 3)/(''k'' - 1)) / ((''k'' + 2)/''k'')<sup>2</sup>:


: = k<sup>2</sup>/(k+1)<sup>3</sup>/(k-1) * (k + 2)k
<math>
\begin {align}
{\rm S}k/{\rm S}(k + 2) &= \frac {k^2/(k^2 - 1)}{(k + 2)^2 / ((k + 2)^2 - 1)} \\
&= \frac {k^2}{(k + 1)(k - 1)} \cdot \frac{(k + 3)(k + 1)}{(k + 2)^2} \\
&= \frac {k^2}{k - 1} \cdot \frac {k + 3}{(k + 2)^2} \\
&= \frac {k + 3}{k - 1} \cdot \frac{k^2}{(k + 2)^2} \\
&= \frac {(k + 3)/(k - 1)}{((k + 2)/k)^2}
\end {align}
</math>


: = k<sup>3</sup>/(k+1)<sup>3</sup> * (k + 2)/(k - 1) = ((k+2)/(k-1)) / ((k+1)/k)<sup>3</sup>
For semiparticulars, we also want to show that S''k''/S(''k'' + 2) is superparticular for all but the case of S(4''n'' - 1)/S(4''n'' + 1) which is odd-particular:


For semiparticulars, we want to show that Sk/S(k+2) = ((k+3)/(k-1)) / ((k+2)/k)<sup>2</sup>:
<math>
\begin {align}
{\rm S}k/{\rm S}(k + 2) &= \frac {k + 3}{k - 1} \cdot \frac{k^2}{(k + 2)^2} \\
&= \frac {k^3 + 3k^2}{(k - 1)(k^2 + 4k + 4)} \\
&= \frac {k^3 + 3k^2}{k^3 + 4k^2 + 4k - k^2 - 4k - 4} \\
&= \frac {k^3 + 3k^2}{k^3 + 3k^2 - 4}
\end {align}
</math>


Sk/S(k+2) = ( k<sup>2</sup>/(k<sup>2</sup> - 1) )/( (k+2)<sup>2</sup>/((k+2)<sup>2</sup> - 1) )
This result will be useful, so we will refer to it as '''[Eq. 1]''': S''k''/S(''k'' + 2) = (k<sup>3</sup> + 3k<sup>2</sup>)/(k<sup>3</sup> + 3k<sup>2</sup> - 4)


: = k<sup>2</sup>/(k+1)/(k-1) * (k+3)(k+1)/(k+2)<sup>2</sup>
: Note that when ''k'' = 2''n'' in [Eq. 1], everything in the numerator and denominator is divisible by 4 because the only instances of ''k'' have it raised to a power of 2 or greater meaning there will be a factor of (2''n'')<sup>2</sup> = 4''n''<sup>2</sup>, therefore S''k''/S(''k'' + 2) is superparticular when ''k'' is even.


: = k<sup>2</sup>/(k-1) * (k+3)/(k+2)<sup>2</sup>
: When ''k'' = 4''n'' + 1 in [Eq. 1], we have to do some extra work to show the result is superparticular:


: = (k+3)/(k-1) * k<sup>2</sup> / (k+2)<sup>2</sup> = ((k+3)/(k-1)) / ((k+2)/k)<sup>2</sup>
: (4''n'' + 1)<sup>3</sup> = (4''n'')<sup>3</sup> + 3·(4''n'')<sup>2</sup> + 3·4''n'' + 1 is of the form 4''m'' + 1.


For semiparticulars, we also want to show that Sk/S(k+2) is superparticular for all but the case of S(4n-1)/S(4n+1) which is odd-particular:
: (4''n'' + 1)<sup>2</sup> = (4''n'')<sup>2</sup> + 2·4''n'' + 1 is also of the form 4m+1.


Sk/S(k+2) = (k+3)/(k-1) * k<sup>2</sup> / (k+2)<sup>2</sup>
: Therefore we can replace their occurrences in [Eq. 1] with 4''m'' + 1 and 4''a'' + 1 respectively, without having to worry about what ''m'' and ''a'' are (as we only need to know that ''m'' and ''a'' are positive integers). Therefore to show S(4''n'' + 1)/S(4''n'' + 3) is superparticular, we set k = 4''n'' + 1 in [Eq. 1] and then do the replacements and simplify to a superparticular:


: = (k<sup>3</sup> + 3k<sup>2</sup>)/( (k-1)(k<sup>2</sup> + 4k + 4) )
: <math>
\begin {align}
{\rm S}(4n + 1)/{\rm S}(4n + 3) &= \frac {(4n + 1)^3 + 3(4n + 1)^2}{(4n + 1)^3 + 3(4n + 1)^2 - 4} \\
&= \frac {(4m + 1) + 3(4a + 1)}{(4m + 1) + 3(4a + 1) - 4} \\
&= \frac {4m + 4(3a) + 4}{4m + 4(3a)} \\
&= \frac {m + 3a + 1}{m + 3a}
\end {align}
</math>


: = (k<sup>3</sup> + 3k<sup>2</sup>)/( k<sup>3</sup> + 4k<sup>2</sup> + 4k - k<sup>2</sup> - 4k - 4 )
: Then for the final case we want to show that S(4''n'' - 1)/S(4''n'' + 1) is odd-particular by setting k = 4''n'' - 1 in [Eq. 1]:


: = (k<sup>3</sup> + 3k<sup>2</sup>)/( k<sup>3</sup> + 3k<sup>2</sup> - 4 )
: S(4''n'' - 1)/S(4''n'' + 1) = ( (4''n'' - 1)<sup>3</sup> + 3(4''n'' - 1)<sup>2</sup> )/( (4''n'' - 1)<sup>3</sup> + 3(4''n'' - 1)<sup>2</sup> - 4 )
 
This result will be useful, so we will refer to it as '''[Eq. 1]''': Sk/S(k+2) = (k<sup>3</sup> + 3k<sup>2</sup>)/( k<sup>3</sup> + 3k<sup>2</sup> - 4 )
 
: Note that when k = 2n in [Eq. 1], everything in the numerator and denominator is divisible by 4 because the only instances of k have it raised to a power of 2 or greater meaning there will be a factor of (2n)<sup>2</sup> = 4n<sup>2</sup>, therefore Sk/S(k+2) is superparticular when k is even.
 
: When k = 4n+1 in [Eq. 1], we have to do some extra work to show the result is superparticular:
 
: (4n+1)<sup>3</sup> = (4n)<sup>3</sup> + 3*(4n)<sup>2</sup> + 3*4n + 1 is of the form 4m+1.
 
: (4n+1)<sup>2</sup> = (4n)<sup>2</sup> + 2*4n + 1 is also of the form 4m+1.
 
: Therefore we can replace their occurrences in [Eq. 1] with 4m+1 and 4a+1 respectively, without having to worry about what m and a are (as we only need to know that m and a are positive integers). Therefore to show S(4n+1)/S(4n+3) is superparticular, we set k=4n+1 in [Eq. 1] and then do the replacements and simplify to a superparticular:
 
: S(4n+1)/S(4n+3) = ( (4n+1)<sup>3</sup> + 3(4n+1)<sup>2</sup> )/( (4n+1)<sup>3</sup> + 3(4n+1)<sup>2</sup> - 4 )
 
:: = ( 4m+1 + 3(4a+1) )/( 4m+1 + 3(4a+1) - 4)
 
:: = ( 4m + 4(3a) + 4 )/( 4m + 4(3a) ) = ( m + 3a + 1 )/( m + 3a )
 
Then for the final case we want to show that S(4n-1)/S(4n+1) is odd-particular by setting k=4n-1 in [Eq. 1]:
 
S(4n-1)/S(4n+1) = ( (4n-1)<sup>3</sup> + 3(4n-1)<sup>2</sup> )/( (4n-1)<sup>3</sup> + 3(4n-1)<sup>2</sup> - 4 )


: As before we make replacements:
: As before we make replacements:


: (4n-1)<sup>3</sup> = (4n)<sup>3</sup> - 3*(4n)<sup>2</sup> + 3*4n - 1 is of the form 4m-1 so will be replaced with such.
: (4''n'' - 1)<sup>3</sup> = (4''n'')<sup>3</sup> - 3·(4''n'')<sup>2</sup> + 3·4''n'' - 1 is of the form 4''m'' - 1 so will be replaced with such.


: (4n+1)<sup>2</sup> = (4n)<sup>2</sup> - 2*4n + 1 is of the form 4a+1 so will be replaced with such.
: (4''n'' + 1)<sup>2</sup> = (4''n'')<sup>2</sup> - 2·4''n'' + 1 is of the form 4''a'' + 1 so will be replaced with such.


: Therefore:
: Therefore:


: S(4n-1)/S(4n+1) = ( 4m-1 + 3(4a+1) )/( 4m-1 + 3(4a+1) - 4 )
: <math>
 
\begin {align}
:: = ( 4m-1 + 4(3a) + 3 )/( 4m-1 + 4(3a) + 3 - 4 )
{\rm S}(4n - 1)/{\rm S}(4n + 1) &= \frac {(4n - 1)^3 + 3(4n - 1)^2}{(4n - 1)^3 + 3(4n - 1)^2 - 4} \\
 
&= \frac {(4m - 1) + 3(4a + 1)}{(4m - 1) + 3(4a + 1) - 4} \\
:: = ( 4m + 4(3a) + 2 )/( 4m + 4(3a) - 2 )
&= \frac {4m + 4(3a) + 2}{4m + 4(3a) - 2} \\
 
&= \frac {2m + 2(3a) + 1}{2m + 2(3a) - 1}
:: = ( 2m + 2(3a) + 1 )/( 2m + 2(3a) - 1 )
\end {align}
</math>


:: ...which is of the form (2x + 1)/(2x - 1) meaning it is odd-particular.
: … which is of the form (2''x'' + 1)/(2''x'' - 1) meaning it is odd-particular.


In conclusion: Sk/S(k+2) is superparticular for k =/= 3 (mod 4) and is odd-particular when k = 3 (mod 4).
In conclusion: S''k''/S(''k'' + 2) is superparticular for k ≠ 3 (mod 4) and is odd-particular when k = 3 (mod 4).


[[Category:Elementary math]]
[[Category:Elementary math]]
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