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.

Answer:

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.