## 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{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}
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.