Question

show that p2 will leave a remainder 1 when divided by 8 if p is a positive odd integer

Answer 1

p = 2n + 1 where n = 0, 1, 2, 3, 4, ... 
=> p² = 4n² + 4n + 1 
= 4(n² + n) + 1 
= 4n(n+1) + 1 
but n(n+1) is even because if n is odd, n+1 is even - otherwise n is even 
therefore 4n(n+1) + 1 = 8m + 1 where m is an integer. 
=> p² = 1 
hope this will help u:)

Answer 2

As p is odd
p= 2n+1
p²= 4n²+1+4n

now p² becomes 4n(n+1)+1
now n+1 is odd whr n is even.
thus,
n(n+1) will be even

now taking n=2m

p²= 8m(2m+1)+1

hence we get remainder 1 if p is positive odd integer on dividing it by 8