Question

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

Answer 1

Any odd number can be expressed as 2a+1, for some integer a. We have

 

(2a+1)2 = 4a2 + 4a + 1 (mod 8)

    = 4a(a+1) + 1 (mod 8)

 

But, we know if a is even then a+1 is odd, or vice versa. The point is a(a+1) must be even, since one of two consecutive integers must be even. Thus, we can expression a(a+1) as 2b, for some integer b. Now we have:

 

   =  4*2b + 1 (mod 8)

   =  8b + 1 (mod 8)

   =1 (mod 8)

Hence Proved