Question

Use MI to prove that \(n^2-1\) is divisible by 8 for any positive odd integer n.

Answer 1

Attempt:
Let p(n) be the statement \(n^2−1\) is divisible by 8
When n = 1
p(1) = 1^2 - 1 = 0, which is divisible by 8.
So, p(1) is true.
Suppose p(k) is true for some positive odd integers k, i.e. k^2-1 = 8m, where \(m\in\mathbb{Z^+}\)
Consider p(k+2)
\[(k+2)^2 - 1\]\[=k^2+4k+4-1\]\[=8m+1+4k+4-1\]\[=8m+4(k+1)\]

Then, I don't know how I should proceed from here.

Answer 2

P(n)=n^2-1,where n is an odd integer. (1)
Let n=2k+1,where k is a positive integer.
P(n)=〖(2k+1)〗^2-1 ….(2)
P(1)=3^2-1=8,which is divisible by 8
Assume that p(m) is true.
P(m)=〖(2m+1)〗^2-1 is divisible by 8
Therefore let (2m+1)^2-1=8l
(2m+1)^2=8l+1
P(m+1)={2(m+1)+1}^2-1={(2m+1)+2}^2-1
=(2m+1)^2+4+2*2(2m+1)-1
=(2m+1)^2+4+8m+4-1
=8l+1+8+8m-1
=8 (l+m+1)
=8*an integer
Hence (2) is true for all k
Or (1) is true for all positive odd integers.

Answer 3

"Let n=2k+1,where k is a positive integer"
What about the case when n=1?

Answer 4

let m^2-1=8l ,where l is an odd integer.
m^2=8l+1
Next odd integer=m+2
P(m+2)=(m+2)^2-1=m^2+4m+4-1
=m^2 +4(m+1)-1
=8l+1+4(m+1)-1
=8l+4(m+1)
=8l+8t, m is odd so m+1 is even ,or m+1=2t (say)
=8(l+t)
=8 * an integer
Hence p(m+2) is true.
Hence by induction p(n) is true for all positive integers

Answer 5

correction i should have written k=0,1,2,3,............
and in the beginning p(0)

Answer 6

Ah! I am stupid :\
4(m+1) = 8t is the key!

Answer 7

Thanks :)

Answer 8

yw