Show that p^2 will leave remainder 1 when divided by 8, if p is an odd positive integer.
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exceptionalmeno
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:
Answers & Comments
(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
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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# NIKKY