Question

Show that n^2-1 is divisible by 2, if n is an odd positive integer.​

Answer 1

Answer:

n^2-1= (n-1)(n+1) if n is odd then n-1 and n+1 both are even number hence these are divisible by 2

Answer 2

Method 1

Let n=2k-1

$(2k-1)^2-1=4k^2-4k=2(2k^2-2k)$

∴Divisible by 2

Method 2

(Congruent Equation)

Since n²-1 is (n+1)×(n-1)

And n≡1 (mod 2)

∴n-1≡0 (mod 2)

∴n+1≡0 (mod 2)

∴n²-1≡0 (mod 2)