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<DIV>Here's a simpler proof for this part of the result. Start with the Pythagorean triple 9+16=25. These are all non-zero modulo any prime greater than 5, so multiply by any non-square residue to get a non-square as the sum of two non-squares.</DIV>
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<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV>
<DIV> </DIV> <BR>-----Original Message-----<BR>From: franktaw@netscape.net<BR>To: jens@voss-ahrensburg.de; seqfan@ext.jussieu.fr<BR>Sent: Tue, 17 Jan 2006 11:43:16 -0500<BR>Subject: Re: Numbers with anticlosed sets of non-squares<BR><BR>
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<DIV>There is a well known result that, modulo any odd prime, every residue is the sum of two squares. (The proof is fairly simple, using the pigeonhole principle.) For p > 5, we can show that every residue is the sum of two non-zero squares. If a + b = 1 with a and b non-zero squares, and r is any non-square, then ra + rb = r is a non-square as the sum of two non-squares.</DIV></DIV></DIV></DIV></DIV></DIV>
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