Question

Prove the following statement.

The square of any odd integer has the form 8m+1 for some integer m.

Answer 1

You can prove this statement as follows:

Explanation:

An odd integer is a number of the form
`2k+1` where
`k\in \mathbb{Z}`. Consider the following cases.

Case 1. If
`k` is even we have:
`(2k+1)^(2)=(2(2s)+1)^(2)=(4s+1)^(2)=16s^2+8s+1=8(2s^2+s)+1`.

If we denote by
`m=2s^2+2` we have that
`(2k+1)^(2)=8m+1`.

Case 2. if
`k` is odd we have:
`(2k+1)^(2)=(2(2s+1)+1)^(2)=(4s+3)^(2)=16s^2+24s+9=16s^(2)+24s+8+1=8(2s^(2)+3s+1)+1`.

If we denote by
`m=2s^(2)+24s+1` we have that
`(2k+1)^(2)=8m+1`

This result says that the remainder when we divide the square of any odd integer by 8 is 1.