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.
Table of Contents
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:
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.
