## Proof That The Difference oF the Squares of Bumbers of Form 6n+1 is Divisible by 24

We can start to prove that the difference of the squares of any two numbers of the form\[6m+1\]

is divisible by 24 by first factorising it.\[\begin{equation} \begin{aligned} (6m+1)^2-(6n+1)^2 &=(36m^2+12m+1)-(36n^2+12n+1) \\ &= 36m^2+12m-36n^2-12n \\ &= 36m^2-36n^2+12m-12n \\ &= 36(m-n)(m+n) +12(m-n) \\ &= 12(m-n)(3m+3n+1) \end{aligned} \end{equation}\]

Obviously \[(6m+1)^2-(6n+1)^2\]

is divisible by 12. If we can show \[(m-n)(3m+3n+1) \]

is divisible by 2 we will have proved it is divisible by 24.The table shows all the possible options.

\[m, \; n\] | \[m-n\] | \[3m+3n+1\] | \[(m-n)(3m_3n+1)\] |

\[m\] even, \[n\] even | even | odd | even |

\[m\] odd, \[n\] even | even | odd | even |

\[m\] even, \[n\] odd | odd | even | even |

\[m\] odd, \[n\] even | odd | even | even |

\[(6m+1)^2-(6n+1)^2\]

is divisible by 24.