Prove: b^2 Congruent to 1 (mod 24) When b Coprime to 6

[Fairy111]

Homework Statement

I need to prove that if b is coprime to 6, then b^2 is congruent to 1 (mod 24)

Homework Equations

The Attempt at a Solution

I know that this means the gcd(b,6)=1

Any help though to start this question off would be great thanks.

 

[HallsofIvy]

And gcd(b, 6)= 1 means b must not have 2 or 3 as factors.

 

[JCVD]

If gcd(b,6) = 1 then b = r (mod 6) for only two values of r between 0 and 5. Then we can write b = 6*n + r for some integer n and look at the expansion of (6*n + r)^2.