Prove for any positive odd integer $n$, $3\mid n$ or $n^2 \equiv 1 \pmod {12}$

Question

I try to use $~n=2x+1~$ to prove it and $~n^2-1=4x(x+1)~$. I do not know how to prove it.

Really need help here.

Answer 1

When $n$ is $1$ $ mod$ $ 2$ then $n^2$ is $1$ $ mod$ $ 4$. If $n$ is not divisible by $3$ then $n^2$ is $1$ $ mod$ $ 3$

Thus when $n$ is not divisible by $3$ one has that

$n^2 \equiv 1$ $ mod$ $ 4$

and

$n^2 \equiv 1$ $ mod$ $ 3$

Now since $(4,3) = 1$ one can apply chinese remainder theorem to deduce $n^2 \equiv 1$ $ mod$ $ 12$

Answer 2

For odd n, not divisible by 3, then n = 6k ± 1:

$n^2 = (6k ± 1)^2 = 36k^2 ± 12k + 1 = 12k(3k ± 1) + 1$

Answer 3

There can be 6 possibility of the numbers , $$ 6n , 6n ± 1, 6n±2,6n+3 $$ => Now if the number is of the form : $$ 6n \ or \ 6n+3 $$ it is divisible by 3 so we are done.

=> If the numbers is $$\ 6n±2 $$ then it is an even numbers and so out of the boundary of the Quest.

=> Now if the number is $$6n±1$$ then its square would be $$ 36n ± 12n +1$$ which leaves remainder 1 when divides by 12 .

Hence Proved