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

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The discussion centers on proving the modular relation b ≡ 1 (mod 2) implies b² ≡ 1 (mod 8). The proof begins with the definition of b as b = 1 + 2k for some integer k, leading to the expression b² = 4k² + 4k + 1. Two cases are analyzed: when k is even (k = 2n) and when k is odd (k = 2n + 1), both of which confirm that b² ≡ 1 (mod 8) holds true.

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John112
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\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?
 
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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?
 

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