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Topic: Prove K^2 mod 4 = 1
Replies: 2   Last Post: Jan 23, 2013 4:51 PM

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 johnykeets Posts: 8 From: usa Registered: 1/1/13
Re: Prove K^2 mod 4 = 1
Posted: Jan 9, 2013 12:28 AM

One of the procedures for solving this is follows:
Let k = 2n + 1 where n is an integer (in Z).
Then k^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 4(n^2 + n) + 1 = 4m + 1 for some integer m (m in Z).
By the definition of congruence something is congruent z = r (MOD q) iff z = pq + r where 0 <= r < q for some integers p, r, q and z.
By letting z = k^2, p = m, r = 1, and q = 4 and noting 0 <= 1 < 4 and all variables are integers, we get our condition that
k^2 = (1 MOD 4)
Hence Prooved.

Message was edited by: johnykeets

Message was edited by: johnykeets

Date Subject Author
12/14/12 brian jasper
12/22/12 grei
1/23/13 johnykeets