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?