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        <item>
            <title>Re: Prove K^2 mod 4 = 1</title>
        
            <link>http://mathforum.org/kb/thread.jspa?messageID=8023862&amp;tstart=0#8023862</link>

        

            <description><![CDATA[One of the procedures for solving this is follows:<br>Let k = 2n + 1 where n is an integer (in Z).<br>Then k^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 4(n^2 +]]></description>

        

            <jf:creationDate>Jan 9, 2013 12:28:58 AM</jf:creationDate>
            <jf:modificationDate>Jan 23, 2013 4:51:24 PM</jf:modificationDate>
            <jf:author>johnykeets</jf:author>
            <jf:replyCount>0</jf:replyCount>
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        <item>
            <title>Re: Prove K^2 mod 4 = 1</title>
        
            <link>http://mathforum.org/kb/thread.jspa?messageID=7942564&amp;tstart=0#7942564</link>

        

            <description><![CDATA[I would suggest factoring the "4" out of the first two terms- 4n^2+ 4n+ 1= 4(n^2+ n)+ 1- to make it obvious.<br>]]></description>

        

            <jf:creationDate>Dec 22, 2012 7:45:09 AM</jf:creationDate>
            <jf:modificationDate>Dec 22, 2012 7:45:09 AM</jf:modificationDate>
            <jf:author>GuyinMississippi</jf:author>
            <jf:replyCount>0</jf:replyCount>
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        <item>
            <title>Prove K^2 mod 4 = 1</title>
        
            <link>http://mathforum.org/kb/thread.jspa?messageID=7936890&amp;tstart=0#7936890</link>

        

            <description><![CDATA[I have been working on this for far too long:(<br><br>Can someone let me know if this is correct?<br><br>"k = 2n + 1. Then k^2 (is congruent to) 4n^2]]></description>

        

            <jf:creationDate>Dec 14, 2012 7:04:00 PM</jf:creationDate>
            <jf:modificationDate>Dec 14, 2012 7:04:00 PM</jf:modificationDate>
            <jf:author>brianmans</jf:author>
            <jf:replyCount>2</jf:replyCount>
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