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
Since the product is always even  
\[(6m+1)^2-(6n+1)^2\]
  is divisible by 24.