Proving Modular Relations: b ≡ 1 (mod 2) ⇒ b² ≡ 1 (mod 8)

[John112]

\forall b\in Z b \equiv 1 (mod 2) \Rightarrow b^{2} \equiv 1 (mod 8)

How do I go about proving this? Can the Chinese Remainder Theorem be used to prove this or is there something easier?

 

[HallsofIvy]

Directly from the definition, b\conguent 1 (mod 2) means that b= 1+ 2k[/tex] for some integer k. Then b^2= 4k^2+ 4k+ 1= 4(k^2+ k)+ 1

Now consider two cases:
1) k is even: k= 2n for some integer n. What is 4(k^2+ k)+ 1 in this case?

2) k is odd: k= 2n+ 1 for some integer n. What is 4(k^2+ k)+ 1 in this case?