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 m^{2} =8n+1 for some integer n.

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## 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}

= 16k^{2}+8k+1

= 8(2k^{2}+k)+1

= 8n+1

where n=2k^{2}+k is an integer.

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

Now, (4k+3)^{2}

= 16k^{2}+24k+9

= 8(2k^{2}+3k+1)+1

= 8n+1

where n=2k^{2}+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:**

## Examples

**Example1:** Note that 3 is an odd integer. Its square 3^{2}=9 = 8⋅1+1, so it is of the form 8n+1 where n=1.

**Example2:** The square of the odd integer 5 is 5^{2}=25 = 8⋅3+1, so it is of the form 8n+1 where n=3.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.