Question

Show that for odd positive integer to be a perfect square, it should be of the form 8k+1.

Answer 1

Hello.

If an odd number is a perfect square it must be of the form 
(2n±1)² 
= 4n²+4n+1 

If n is even: 
4n² is divisible by 8 and 4n is divisible by 8 so it is of the form 8k+1 

If n is odd 
4n² = 4(2k+1)² = 16k²+16k+4 
4n = 4(2k+1) = 8k + 4 

4n²+4n+1 
= 16k²+24k+9 
= 16k²+24k+8+1 
= 8(2k²+3k+1)+1 
which is of the form 8k+1 

If n is even (2n±1)² must be of the form 8k+1 
If n is odd (2n±1)² must be of the form 8k+1 

(2n±1)² must be of the form 8k+1 for all integer values of n 

The square of any odd number must be of the form 8k+1.

Hope it helps.