Question

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

Answer 1

Step-by-step explanation:

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

gudspeling · 10 years ago