# Square of an odd integer is of the form 8n+1 [Proof]

The square of an odd integer is of the form 8n+1 where n is an integer. That is, If m is an odd integer then m2 =8n+1 for some integer n.

## Prove that the square of an odd integer is of the form 8n+1.

By division algorithm, when we divide an integer by 4 the remainder will be either 0, 1, 2, or 3. Therefore, any integer can be written as one of the forms:

4k, 4k+1, 4k+2, 4k+3

where k is an integer.

Among them the odd integers are 4k+1, 4k+3.

Case1: Odd integer is of the form 4k+1.

Now, (4k+1)2

= 16k2+8k+1

= 8(2k2+k)+1

= 8n+1

where n=2k2+k is an integer.

Case1: Odd integer is of the form 4k+3.

Now, (4k+3)2

= 16k2+24k+9

= 8(2k2+3k+1)+1

= 8n+1

where n=2k2+3k+1 is an integer.

This proves that the square of an odd integer is of the form 8n+1 for some integer n.

Related Topic:

What are odd integers?

What are even integers?

## Examples

Example1: Note that 3 is an odd integer. Its square 32=9 = 8⋅1+1, so it is of the form 8n+1 where n=1.

Example2: The square of the odd integer 5 is 52=25 = 8⋅3+1, so it is of the form 8n+1 where n=3.

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