When the square of any odd number, greater than 1, is divided by 8, it always leaves remainder 1 (b) 6 (c) 8 (d) Cannot be determined

Explanation

Let `P(n):(2r+1)^(2n), forall n in N and r in I`.
Step I For `n=1`.
`P(1):(2r+1)^2=4r^2+4r+1=4r(r+1)+1=8p+1,p in I " "[because r(r+1)"is an even integer"]`
Therefore , `P(1)` is true ,
Step II Assume P(n) is true for n=k , then
`P(k):(2r+1)^2k` is divisible by 8 levaes remainder 1.
`rArr P(k)=8m+1,n in I`, where m is a positive integer .
Step III For `n=k+1`. brgt `therefore P(k)=(2r+1)2(k+1)`
`=(2r+1)^(2k)(2r+1)^2`
`=(8m+1)(8p+1)`
`64mp+8(m+p)+1`
`=8(8mp+m+p)+1`
which is true for `n=k+1` as `8mp+m+p` is an integer. Hence , by the principle of mathematical induction, when P(n) is divided by 8 leaves the ramainder 1 for all `n in N`.