Question

Pprove that $n^{2}$ - 1 is divisible by 16 for all odd integers n.

Answer 1

Sol: If 'n' is an odd positive integer, show that (n2-1) Is divisible by 8?? Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p. Let n = 4p+ 1, (n2 – 1) = (4p + 1)2 – 1 = 16p2 + 8p + 1 = 16p2 + 8p = 8p (2p + 1) ⇒ (n2 – 1) is divisible by 8. (n2 – 1) = (4p + 3)2 – 1 = 16p2 + 24p + 9 – 1 = 16p2 + 24p + 8 = 8(2p2 + 3p + 1) ⇒ n2– 1 is divisible by 8. Therefore, n2– 1 is divisible by 8 if n is an odd positive integer.

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Answer 2

Answer:

Step-by-step explanation:

if n = 3 then n^2 - 1 = 8 and 8 is not divisible by 16

if n = 5 then n^2 - 1 = 24 and 24 is not divisible by 16.

the question is therefore not right.

However for odd integers n^2 - 1 is Always divisible by 8.