Question

if n is any odd integer then

show that n^2 -1 is divisble by 8

Answer 1

let n be odd number, then we can write \(n= 2k + 1\) for some integer \(k\)

now,
\[n^{2} - 1 = ....\]

change n with 2k + 1

Answer 2

we can't take 4k+1

Answer 3

be careful...,

\[n = 2k + 1\]

\[n^{2} -1 = (2k+1)^2 -1\]

can you continue it?

Answer 4

yes
i done and solved it also

Answer 5

2(2n)+1 is also, any odd number. which would be useful

Answer 6

if you can simplify it to 8(....) then you know its divisible by 8