Question

Show that p^2 will leave 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

Answer 2

HELLO DEAR ,

HERE IS YOUR ANSWER,

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Given that p is any positive odd integer. 
Therefore, p is of the form:
p = 2q + 1 where q is any positive integer.  

p2 = 4q2 + 4q + 1 
p2 = 4q (q+1) + 1

q(q+1) is a even integer as the product of any two consecutive integers is even. Therefore q(q+1) is a multiple of 2 and let it be 2n

Therefore,
p2 = 4 × 2n + 1 
p2 = 8n + 1 

From Euclid's Lemma we can say that the remainder is 1 when p2 is divided by 8. 

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HOPE IT HELPS U :))

# NIKKY