Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Math Topics » discretemath

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
johnykeets

Posts: 8
From: usa
Registered: 1/1/13
Re: Prove K^2 mod 4 = 1
Posted: Jan 9, 2013 12:28 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.