Show that [tex]$n^2 - 1$[/tex] is divisible by 8 if [tex]$n$[/tex] is an odd positive integer.

n − an o dd p os i t i v e in t e g er ⇒ n = 2 k + 1 an d k ∈ N a t u r a l 2 − 1 = ( 2 k + 1 ) 2 − 1 = ( 2 k ) 2 + 2 ⋅ 2 k ⋅ 1 + 1 2 − 1 = 4 k 2 + 4 k = = 4 k ( k + 1 ) i f k − o dd t h e n ( k + 1 ) − e v e n , t h e n k ( k + 1 ) − e v e n ⇒ k ( k + 1 ) = 2 m an d m ∈ N ⇒ 4 k ( k + 1 ) = 4 ⋅ 2 m = 8 m i f k − e v e n , t h e n k ( k + 1 ) − e v e n ⇒ k ( k + 1 ) = 2 p an d p ∈ N ⇒ 4 k ( k + 1 ) = 4 ⋅ 2 p = 8 p
− − − − − − − − − − − − − − − − − − − − − − − − f or e a c h k ∈ N t h e p ro d u c t o f k ( k + 1 ) i s e v e n , so : i f ′ n ′ i s an o dd p os i t i v e in t e g er , t h e n n 2 − 1 i s d i v i s ib l e 8.

Answered by kate200468 | 2024-06-10

We showed that if n is an odd positive integer, then n 2 − 1 can be expressed as 8 m , meaning it is divisible by 8. This is because the expression simplifies to 4 k ( k + 1 ) , where k ( k + 1 ) is always even. Therefore, n 2 − 1 is divisible by 8.
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Answered by kate200468 | 2025-05-13