S-expression: Difference between revisions
m →Mathematical derivation: correction |
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: (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. | : (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. | ||
: (4''n'' + 1)<sup>2</sup> = (4''n'')<sup>2</sup> + 2·4''n'' + 1 is also of the form | : (4''n'' + 1)<sup>2</sup> = (4''n'')<sup>2</sup> + 2·4''n'' + 1 is also of the form 4''m'' + 1. | ||
: 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: | : 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: | ||