Question

(ii) Let a be a number of the form a = 6n + 1 or a = 6n−1 for some integer n. Show that a2 ≡ 1 (mod 24). (Hint: Calculate (6n±1)2 and consider two different cases for n.)

Answer 1

If $a=6n+1$ then
a = 36n2 + 12n + 1
. If $a=6n-1$ then $a^2=36n^2-12n+1$ . Now, there are two cases:

Case I: is even, say $n=2m$ ; then either

$a^2=36n^2+12n+1=144m^2+24m+1=24(6m^2+m)+1$

or

$a^2=36n^2-12n+1=144m^2-24m+1=24(6m^2-m)+1$

Thus, either way we have $a^2\equiv 1\mod 24$ .

Case II: is odd, say $n=2m+1$ ; then either

$\begin{align*}a^2&=36n^2+12n+1\\ &=144m^2+144m+36+24m+12+1\\ &=24(6m^2+7m+2)+1\end{align*}$

or

$\begin{align*}a^2&=36n^2-12n+1\\ &=144m^2+144m+36-24m-12+1\\ &=24(6m^2+5m+1)+1\end{align*}$

Thus, either way we have $a^2\equiv 1\mod 24$ .

This proves that $a^2\equiv 1\mod 24$ ​​​​​​​.