Strongly regular graph: Difference between revisions

Revision as of 22:37, 27 June 2022

Revision as of 02:43, 21 September 2022

Line 174:

If the denominator <math>\sqrt{4k - 3}</math> is an integer ''t'', then <math>4k - 3</math> is a perfect square <math>t^2</math>, so <math>k = \frac{t^2 + 3}{4}</math>. Substituting:

:<math>\begin{align}

:<math>M_{\pm} = \frac{1}{2}\left[\left(\frac{t^2 + 3}{4}\right)^2 \pm \frac{\frac{t^2 + 3}{2} - \left(\frac{t^2 + 3}{4}\right)^2}{t}\right]</math>

:<math>\implies 32 M_{\pm} &= \frac{1}{2} \left[\left(\frac{t^2 + 3}{4}\right)^2 \pm \frac{8(\frac{t^2 + 3)}{2} - \left(\frac{t^2 + 3}{4}\right)^2}{t}</math>\right] \\

:<math>\implies 32 M_{\pm} &= (t^42 + 6t3)^2 + 9 \pm \frac{- 8(t^42 + 2t3) - (t^2 + 153)^2}{t}</math> \\

:<math>\implies 32 M_{\pm} &= t^4 + 6t^2 + 9 \pm \left(frac{- t^34 + 2t^2 + \frac{15}{t} \\right)</math>

&= t^4 + 6t^2 + 9 \pm \left(-t^3 + 2t + \frac{15}{t}\right)

\end{align}</math>

Since both sides are integers, <math>\frac{15}{t}</math> must be an integer, therefore ''t'' is a factor of 15, namely <math>t \in \{\pm 1, \pm 3, \pm 5, \pm 15\}</math>, therefore <math>k \in \{1, 3, 7, 57\}</math>. In turn: